
Chapter 30 – Gauss’s Law
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Question 1:
A small plane area is rotated in an electric field. In which orientation of the area, is the flux of the electric field through the area maximum? In which orientation is it zero?
Answer:
The flux of an electric field
E→through a surface area
∆S→is given by
∆ϕ=E→.∆S→, where
∆ϕis the flux. Therefore,
∆ϕ=E∆S Cosθ. Here,
θis the angle between the electric field
E→and the normal to the surface area.
Thus, for the flux to be maximum, cos
θshould be maximum. Thus, for
θ= 0, the flux is maximum, i.e. the electric field lines are perpendicular to the surface area.
The flux is minimum if
θ= 90. Thus, cos
θ= 0 and, hence, flux is also 0. Thus, if the electric field lines are parallel to the surface area, the flux is minimum.
Question 2:
A circular ring of radius r made of a nonconducting material is placed with its axis parallel to a uniform electric field. The ring is rotated about a diameter through 180°. Does the flux of the electric field change? If yes, does it decrease or increase?
Answer:
It is given that the circular ring, made of a nonconducting material, of radius r is placed with its axis parallel to a uniform electric field.This means that both the electric field and the area vector are parallel to each other (area vector is always perpendicular to the surface area). Thus, the flux through the ring is given by
E→.S→= ES cos 0 = E(
πr2).
Now, when the ring is rotated about its diameter through 180^{0}, the angle between the area vector and the electric field becomes 180^{0}. Thus, the flux becomes
E
(πr2).
Question 3:
A charge Q is uniformly distributed on a spherical shell. What is the field at the centre of the shell? If a point charge is brought close to the shell, will the field at the centre change? Does your answer depend on whether the shell is conducting or nonconducting?
Answer:
The field at the centre of the shell is zero. As all the charge given to a conductor resides on the surface, the field at any point inside the conducting sphere is zero. Also, the charge distribution at the surface is uniform; so all the electric field vectors due to these charges at the centre are equal and opposite. So, they cancel each other, resulting in a zero net value of the field.
When a charge is brought near the shell, due to induction, opposite polarity charges induce on the surface nearer to the charge and the same polarity charges appear on the face farther from the charge. In this way, a field is generated inside the shell. Hence, the field at the centre is nonzero.
Yes, our answer changes in case of a nonconducting spherical shell. As the charge given to the surface of a nonconducting spherical shell spreads nonuniformly, there is a net electric field at the centre of the sphere.
Question 4:
A spherical shell made of plastic, contains a charge Q distributed uniformly over its surface. What is the electric field inside the shell? If the shell is hammered to deshape it, without altering the charge, will the field inside be changed? What happens if the shell is made of a metal?
Answer:
As the shell is made of plastic, it is nonconducting. But as the charge is distributed uniformly over the surface of the shell, the sum of all the electric field vectors at the centre due to this kind of distribution is zero. But when the plastic shell is deformed, the distribution of charge on it becomes nonuniform. In other words, the sum of all the electric field vectors is nonzero now or the electric field exists at the centre now.
In case of a deformed conductor, the field inside is always zero.
Question 5:
A point charge q is placed in a cavity in a metal block. If a charge Q is brought outside the metal, will the charge q feel an electric fore?
Answer:
Yes, the charge Q will feel an electric force, as the charge q given to the metal block appears on the surface. Hence, it exerts an electric force on the charge Q.
Question 6:
A rubber balloon is given a charge Q distributed uniformly over its surface. Is the field inside the balloon zero everywhere if the balloon does not have a spherical surface?
Answer:
No, the field is not zero everywhere, as the electric field vector due to the charge distribution does not cancel out at any place inside the balloon because of its nonspherical shape.
Question 7:
It is said that any charge given to a conductor comes to its surface. Should all the protons come to the surface? Should all the electrons come to the surface? Should all the free electrons come to the surface?
Answer:
Protons never take part in any electrical phenomena because they are inside the nuclei and are not able to interact easily. These are the free electrons that are responsible for all electrical phenomena. So, if a conductor is given a negative charge, the free electrons come to the surface of the conductor. If the conductor is given a positive charge, electrons move away from the surface and leave a positive charge on the surface of the conductor.
Question 1:
A charge Q is uniformly distributed over a large plastic plate. The electric field at a point P close to the centre of the plate is 10 V m^{−1}. If the plastic plate is replaced by a copper plate of the same geometrical dimensions and carrying the same charge Q, the electric field at the point P will become
(a) zero
(b) 5 V m^{−1}
(c) 10 V m^{−1}
(d) 20 V m^{−1}
Answer:
(c) 10 V m^{1}
The electric field remains same for the plastic plate and the copper plate, as both are considered to be infinite plane sheets. So, it does not matter whether the plate is conducting or nonconducting.
The electric field due to both the plates,
E =
σε0
Question 2:
A metallic particle with no net charge is placed near a finite metal plate carrying a positive charge. The electric force on the particle will be
(a) towards the plate
(b) away from the plate
(c) parallel to the plate
(d) zero
Answer:
(a) towards the plate
The particle is a conductor. When it is brought near a positively charged metal plate, opposite charge is induced on its face nearer to the plate, i.e. negative and the same amount of charge, but of opposite polarity, goes to the farther end, i.e. positive.
Now, the attractive force is due to charges of opposite polarity. As they are at a lesser distance than the same polarity charges, the force of attraction is greater than the force of repulsion. In other words, the force on the particle is towards the plate.
Question 3:
A thin, metallic spherical shell contains a charge Q on it. A point charge q is placed at the centre of the shell and another charge q_{1} is placed outside it as shown in the figure (30Q1). All the three charges are positive. The force on the charge at the centre is
(a) toward left
(b) towards right
(c) upward
(d) zero
Figure
Answer:
(d) zero
A charge placed outside a conductor can induce a charge on it or can affect the charge on its surface. But it does not affect what is contained inside the conductor. Similarly, charge q_{1} can affect charge Q; still, the force inside the conductor remains zero. An analogy to the above statement is that when lightning strikes a car, the charge that appears on the car’s metal surface does not affect its interior. Due to this passengers are recommended to sit inside the car.
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Question 4:
Consider the situation of the previous problem. The force on the central charge due to the shell is
(a) towards left
(b) towards right
(c) upward
(d) zero
Answer:
(b) towards right
This question can be answered using the concept of electric field lines.
We know that electric field lines emerge from a positive charge and move towards a negative charge. Now, charge Q, on the face in front of charge q_{1}_{,} tries to nullify the field lines emerging from charge q_{1}_{.} So, the charge on the farther face of the conductor dominates and, hence, a force appears to the right of charge q.
Question 5:
Electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius 10 cm surrounding the total charge is 25 V m. The flux over a concentric sphere of radius 20 cm will be
(a) 25 V m
(b) 50 V m
(c) 100 V m
(d) 200 V m
Answer:
(a) 25 V m
The flux through a surface does not depend on its shape and size; it only depends upon the charge enclosed inside the volume. Here, the charge enclosed by the sphere of radius 10 cm and the sphere of radius 20 cm is same so the flux through them will also be same.
Question 6:
(30Q2a) shows an imaginary cube of edge L/2. A uniformly charged rod of length L moves towards the left at a small but constant speed
ν.At t = 0, the left end just touches the centre of the face of the cube opposite it. Which of the graphs shown in the figure (30Q2b) represents the flux of the electric field through the cube as the rod goes through it?
FigureFigure
Answer:
(d)
At first, when the rod is inserted into the cube, the flux start increasing. When the rod is fully inserted, the flux becomes constant and remains constant for the remaining L/2 length of the rod. After that, as the rod moves out of the cube, the flux starts decreasing. These processes are depicted only by curve (d).
Question 7:
A charge q is placed at the centre of the open end of a cylindrical vessel (figure 30Q3). The flux of the electric field through the surface of the vessel is
(a) zero
(b)
q/εν(c)
q/2εν(d)
2q/εν.
Figure
Answer:
(C) q/2
∈0According to Gauss’s Law, the flux through a closed cylindrical Gaussian surface is q/
∈0. But the question is about an open cylindrical vessel. Now, take another identical vessel and make a closed Gaussian surface enclosing the charge, as shown in the figure.
Total flux linked with the closed Gaussian surface,
ϕT=q∈0Flux linked with the surface of a open ended cylindrical vessel,
ϕ=ϕT2=q2∈0
Question 1:
Mark the correct options:
(a) Gauss’s Law is valid only for symmetrical charge distributions.
(b) Gauss’s Law is valid only for charges placed in vacuum.
(c) The electric field calculated by Gauss’s Law is the field due to the charge inside the Gaussian surface.
(d) The flux of the electric field through a closed surface due to all the charges is equal to the flux due to the charges enclosed by the surface.
Answer:
(d) The flux of the electric field through a closed surface due to all the charges is equal to the flux due to the charges enclosed by the surface.
The contribution of flux on the closed surface due to the charges lying outside the surface is zero because number of field line entering the closed surface is equal to the number of field lines coming out of the surface so the net contribution of the charge lying outside the closed surface to the flux is zero. Therefore, the net flux through the surface due to the charge lying outside the the closed surface is zero. The contribution that counts is only due to the charges lying within the closed surface.
Thus, the flux of the electric field through a closed surface due to all the charges (inside and outside the surface) is equal to the flux due to the charges enclosed by the surface.
Question 2:
A positive point charge Q is brought near an isolated metal cube.
(a) The cube becomes negatively charged.
(b) The cube becomes positively charged.
(c) The interior becomes positively charged and the surface becomes negatively charged.
(d) The interior remains charge free and the surface gets nonuniform charge distribution.
Answer:
(d) The interior remains charge free and the surface gets nonuniform charge distribution.
Since the cube is metallic, the charge gets distributed on the surface and the interior remains chargefree. However, when a positive point charge Q is brought near the metallic cube, a negative charge gets induced on the face near the positive charge Q and an equal positive charge gets induced on the face, which is away from the charge Q. Thus, the metallic surface gets nonuniform charge distribution.
Question 3:
A large nonconducting sheet M is given a uniform charge density. Two uncharged small metal rods A and B are placed near the sheet as shown in the figure (30Q4).
(a) M attracts A.
(b) M attracts B.
(c) A attracts B.
(d) B attracts A.
Figure
Answer:
(a) M attracts A.
(b) M attracts B.
(c) A attracts B.
(d) B attracts A.
Since the nonconducting sheet M is given a uniform charge, it induces a charge in the metal rod A, which further induces a charge in the metal rod B, as shown in the figure. Hence, all the options are correct.
Question 4:
If the flux of the electric field through a closed surface is zero,
(a) the electric field must be zero everywhere on the surface
(b) the electric field may be zero everywhere on the surface
(c) the charge inside the surface must be zero
(d) the charge in the vicinity of the surface must be zero
Answer:
(b) the electric field may be zero everywhere on the surface
(c) the charge inside the surface must be zero
As the flux is zero through the surface, the charge enclosed must be zero. But the electric field is not necessarily zero everywhere on the surface. For example, in the case of a dipole enclosed by the surface, the electric field through the surface is not zero but has some value. So, the correct answers are (b) and (c).
Question 5:
An electric dipole is placed at the centre of a sphere. Mark the correct options.
(a) The flux of the electric field through the sphere is zero.
(b) The electric field is zero at every point of the sphere.
(c) The electric field is not zero anywhere on the sphere.
(d) The electric field is zero on a circle on the sphere.
Answer:
(a) The flux of the electric field through the sphere is zero.
(c) The electric field is not zero anywhere on the sphere.
The sphere encloses a dipole, i.e. two equal and opposite charges. In other words, net charge enclosed in the sphere is zero. Hence, the flux is zero through the sphere.
But the electric field at any point p on the sphere,
E=14π∈0pr33cos2θ+1, where
θis the angle made by the point p with the centre of the dipole.
Hence, we can see that the field is not zero anywhere on the sphere.
Question 6:
Figure (30Q5) shows a charge q placed at the centre of a hemisphere. A second charge Q is placed at one of the positions A, B, C and D. In which position(s) of this second charge, the flux of the electric field through the hemisphere remains unchanged?
(a) A
(b) B
(c) C
(d) D
Figure
Answer:
(a) A
(c) C
These are the only points in the straight line with the charge q and at the brim of the hemisphere. So, field lines emerging from these charges do not affect the flux through the hemisphere due to charge q. On the other hand, the two remaining charges are beside the surface of the hemisphere and not in line with the charge q and not at the brim. So, they will affect the flux. Hence, the correct answers are (a) and (c).
Question 7:
A closed surface S is constructed around a conducting wire connected to a battery and a switch (figure 30Q6). As the switch is closed, the free electrons in the wire start moving along the wire. In any time interval, the number of electrons entering the closed surface S is equal to the number of electrons leaving it. On closing the switch, the flux of the electric field through the closed surface
(a) is increased
(b) is decreased
(c) remains unchanged
(d) remains zero
Figure
Answer:
(c) remains unchanged
(d) remains zero
Initially, there is no charge in the closed surface. As the wire is neutral, the flux initially is zero. Now, if we connect the battery and a current flows through it, the flux remains zero, as the number of electrons entering the surface is equal to number of electrons leaving. That is, net charge enclosed is zero and so is the flux.
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Question 8:
Figure (30Q7) shows a closed surface that intersects a conducting sphere. If a positive charge is placed at point P, the flux of the electric field through the closed surface
(a) will remain zero
(b) will become positive
(c) will become negative
(d) will become undefined
Figure
Answer:
(b) will become positive
A positive charge at point P will induce a negative charge on the near face of the conducting sphere, whereas the positive charge on the farther end of the sphere. As this father end is enclosed or intersected by the closed surface, so the flux through it will become positive due to the induced positive charge.
Question 1:
The electric field in a region is given by
E→=35E0 i→+45 E0 j→ with E0=2·0×103 N C1. Find the flux of this field through a rectangular surface of area 0⋅2 m^{2} parallel to the y–z plane.
Answer:
Given:
Electric field strength,
E→=35E0i^+45 E0j⏜,
where
E0 = 2.0×103 N/CThe plane of the rectangular surface is parallel to the yz plane. The normal to the plane of the rectangular surface is along the x axis. Only
35E0i⏜passes perpendicular to the plane; so, only this component of the field will contribute to flux.
On the other hand,
45E0j⏜moves parallel to the surface.
Surface area of the rectangular surface, a = 0⋅2 m^{2}
Flux,
ϕ=E→.a→ =E×aϕ= 35×2×103×2×101 Nm2/Cϕ=0.24×103 Nm2/Cϕ=240 Nm2/C
Question 2:
A charge Q is uniformly distributed over a rod of length l. Consider a hypothetical cube of edge l with the centre of the cube at one end of the rod. Find the minimum possible flux of the electric field through the entire surface of the cube.
Answer:
Given:
Total charge on the rod = Q
The length of the rod = edge of the hypothetical cube = l
Portion of the rod lying inside the cube, x =
l2
Linear charge density for the rod
=Ql
Using Gauss’s theorem, flux through the hypothetical cube,
Ï• =(Q_{in}/
∈0), where Q_{in} = charge enclosed inside the cube
Here, charge per unit length of the rod =
QlCharge enclosed, Q_{in} =
Ql×l2=Q2Therefore,
Ï•
=Q/2∈0=Q2∈0
Question 3:
Show that there can be no net charge in a region in which the electric field is uniform at all points.
Answer:
It is given that the electric field is uniform. If we consider a surface perpendicular to the electric field, we find that it is an equipotential surface. Hence, if a test charge is introduced on the surface, then work done will be zero in moving the test charge on it.
But if there is some net charge in this region, the test charge introduced on the surface will experience a force due to this charge. This force has a component parallel to the surface; thus, work has to be done in moving this test charge. Thus, the surface cannot be said to be equipotential. This implies that the net charge in the region with uniform electric field is zero.
Question 4:
The electric field in a region is given by
E→=E0 xli→. Find the charge contained inside the cubical volume bound by the surfaces
x=0, x=a, y=0, y=a, z=0 and z=a. Take
E0=5×103 N C1, l=2 cmand a = 1 cm.
Answer:
Given:
Electric field strength,
E→=E0xli^Length,
l=2 cmEdge of the cube,
a = 1 cm
E0 = 5.0 ×103 N/C
It is observed that the flux passes mainly through the surfaces ABCD and EFGH. The surfaces AEFB and CHGD are parallel to the electric field. So, electric flux for these surfaces is zero.
The electric field intensity at the surface EFGH will be zero.
If the charge is inside the cube, then equal flux will pass through the two parallel surfaces ABCD and EFGH. We can calculate flux passing only through one surface. Thus,
E→=E0 xli^At EFGH, x = 0; thus, the electric field at EFGH is zero.
At ABCD, x = a; thus, the electric field at ABCD is
E→=E0 ali^.
The net flux through the whole cube is only through the side ABCD and is given by
ϕ=E→.A→=E0 ali^.ai^=E0 a2l.
Net flux =
ϕ=5×1032×102×1×1022 Nm2/C=25 Nm2/CThus, the net charge,
q = ∈0 ϕ q= 8.85×1012×25 q= 22.125×1013 q= 2.2125×1012 C
Question 5:
A charge Q is placed at the centre of a cube. Find the flux of the electric field through the six surfaces of the cube.
Answer:
According to Gauss’s Law, flux passing through any closed surface is equal to
1∈0times the charge enclosed by that surface.
⇒ϕ=q∈0,
where Ï• is the flux through the closed surface and q is the charge enclosed by that surface.
The charge is placed at the centre of the cube and the electric field is passing through the six surfaces of the cube.vSo, we can say that the total electric flux passes equally through these six surfaces .
Thus, flux through each surface,
ϕ’=Q6 ∈0
Question 6:
A charge Q is placed at a distance a/2 above the centre of a horizontal, square surface of edge a as shown in the figure (30E1). Find the flux of the electric field through the square surface.
Figure
Answer:
Given:
Edge length of the square surface = a
Distance of the charge Q from the square surface = a/2
Area of the plane = a^{2}
Assume that the given surface is one of the faces of the imaginary cube.
Then, the charge is found to be at the centre of the cube.
A charge is placed at a distance of about
a2from the centre of the surface.
The electric field due to this charge is passing through the six surfaces of the cube.
Hence flux through each surface,
ϕ=Q∈0×16=Q6 ∈0Thus, the flux through the given surface is
Q6∈0.
Question 7:
Find the flux of the electric field through a spherical surface of radius R due to a charge of 10^{−7} C at the centre and another equal charge at a point 2R away from the centre (figure 30E2).
Figure
Answer:
Given:
Let charge Q be placed at the centre of the sphere and Q‘ be placed at a distance 2R from the centre.
Magnitude of the two charges = 10^{−7} C
According to Gauss’s Law, the net flux through the given sphere is only due to charge Q that is enclosed by it and not by the charge Q‘ that is lying outside.
So, only the charge located inside the sphere will contribute to the flux passing through the sphere.
Thus,
ϕ=∫E→.ds→=Q∈0=1078.85×1012 ⇒ϕ =1.1×104 Nm2C1
Question 8:
A charge Q is placed at the centre of an imaginary hemispherical surface. Using symmetry arguments and Gauss’s Law, find the flux of the electric field due to this charge through the surface of the hemisphere (figure 30E3).
Figure
Answer:
From Guass’s law, flux through a closed surface,
ϕ=Qen∈0,
where
Q_{en} = charge enclosed by the closed surface
Let us assume that a spherical closed surface in which the charge is enclosed is Q.
The flux through the sphere,
ϕ=Q∈0
Hence for a hemisphere(open bowl), total flux through its curved surface,
ϕ’=Q∈0×12=Q2 ∈0
Question 9:
A spherical volume contains a uniformly distributed charge of density
2·0×104 C m3. Find the electric field at a point inside the volume at a distance 4⋅0 cm from the centre.
Answer:
Given :
Volume charge density, ρ =
2×104 C/m3Let us assume a concentric spherical surface inside the given sphere with radius =
4 cm=4×102 m
The charge enclosed in the spherical surface assumed can be found by multiplying the volume charge density with the volume of the sphere. Thus,
q=ρ×43πr3⇒q=2×104×43πr3The net flux through the spherical surface,
ϕ=q∈0The surface area of the spherical surface of radius r cm:
A =
4πr2Electric field,
E=q∈0×AE= 2×104×4πr3∈0×3×4πr2E= 2×104×r3×∈0The electric field at the point inside the volume at a distance 4⋅0 cm from the centre,
E= (2×104)×(4×102)3×(8.85×1012) N/CE=3.0×105 N/C
Question 10:
The radius of a gold nucleus (Z = 79) is about
7·0×1010 m. Assume that the positive charge is distributed uniformly throughout the nuclear volume. Find the strength of the electric field at (a) the surface of the nucleus and (b) at the middle point of a radius. Remembering that gold is a conductor, is it justified to assume that the positive charge is uniformly distributed over the entire volume of the nucleus and does not come to the outer surface?
Answer:
Given:
Atomic number of gold = 79
Charge on the gold nucleus,
Q=79×(1.6×1019 )CThe charge is distributed across the entire volume. So, using Gauss’s Law, we get:
(a)
ϕ = ∮E→.ds→=Q∈0⇒∮Eds=Q∈0The value of E is fixed for a particular radius.
⇒E∮ds=Q∈0⇒E×4πr2=Q∈0⇒E=Q∈0×4πr2
E = 79×(1.6×1019)(8.85×1012)×4×3.14×(7×1010)2E= 2.315131×1021 N/C(b) To find the electric field at the middle point of the radius:
Radius,
r=72×1010 mVolume,
V= 43πr3= 43×227×3438×1030Net charge
=79×1.6×1019 CVolume charge density
=79×1.6×101943π×343×1030So, the charge enclosed by this imaginary sphere of radius r =3.5
×1010
=79×1.6×101943π×343×1030×43π×3438×1030=79×1.6×10198
⇒ E=79×1.6×10198×4π ∈0.r2 at r = 3.5×1010 =1.16×1021 N/C.As electric charge is given to a conductor, it gets distributed on its surface. But nucleons are bound by the strong force inside the nucleus. Thus, the nuclear charge does not come out and reside on the surface of the conductor. Thus, the charge can be assumed to be uniformly distributed in the entire volume of the nucleus.
Question 11:
A charge Q is distributed uniformly within the material of a hollow sphere of inner and outer radii r_{1} and r_{2} (figure 30E4). Find the electric field at a point P at a distance x away from the centre for r_{1} < x < r. Draw a rough graph showing the electric field as a function of x for 0 < x < 2r_{2} (figure 30E4).
Figure
Answer:
Amount of charge present on the hollow sphere = Q
Inner radii of the hollow sphere =
r1
Outer radii of the hollow sphere =
r2Consider an imaginary sphere of radius x.
The charge on the sphere can be found by multiplying the volume charge density of the hollow spherical volume with the volume of the imaginary sphere of radius (x–
r1).
Charge per unit volume of the hollow sphere,
ρ=Q43πr23r13Charge enclosed by this imaginary sphere of radius x:
q = ρ × Volume of the part consisting of charge
q=43π x3r13 Q43π r23r13q=x3r13r23r13 QAccording to Gauss’s Law,
∮ E.ds=q∈0Here, the surface integral is carried out on the sphere of radius x and q is the charge enclosed by this sphere.
E∮ds=q∈0E(4πx2)=q∈0
E(4π x2) = x3r13Qr23r13 ∈0E = Qx3r134π ∈0 x2 r23r13The electric field is directly proportional to x for r_{1} < x < r.
The electric field for r_{2} < x < 2r_{2,}
E=Q4π∈0x2Thus, the graph can be drawn as:
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Question 12:
A charge Q is placed at the centre of an uncharged, hollow metallic sphere of radius a. (a) Find the surface. (b) If a charge q is put on the sphere, what would be the surface charge densities on the inner and outer surfaces? (c) Find the electric field inside the sphere at a distance x from the centre in the situations (a) and (b).
Answer:
Given:
Amount of charge present at the centre of the hollow sphere = Q
We know that charge given to a hollow sphere will move to its surface.
Due to induction, the charge induced at the inner surface = −Q
Thus, the charge induced on the outer surface = +Q
(a)
Surface charge density is the charge per unit area, i.e.
σ
=chargetotal surface areaSurface charge density of the inner surface,
σin=Q4πa2Surface charge density of the outer surface,
σout=Q4πa2(b)
Now if another charge q is added to the outer surface, all the charge on the metal surface will move to the outer surface. Thus, it will not affect the charge induced on the inner surface. Hence the inner surface charge density,
σin=q4πa2As the charge has been added to the outer surface, the total charge on the outer surface will become (Q+q).
So the outer surface charge density,
σout=q+Q4πa2(c)
To find the electric field inside the sphere at a distance x from the centre in both the situations,let us assume an imaginary sphere inside the hollow sphere at a distance x from the centre.
Applying Gauss’s Law on the surface of this imaginary sphere,we get:
∮ E.ds=Q∈0E∮ ds=Q∈0E(4πx2)=Q∈0 E=Q∈0×14πx2 = Q4π ∈0 x2Here, Q is the charge enclosed by the sphere.
For situation (b):
As the point is inside the sphere, there is no effect of the charge q given to the shell.
Thus, the electric field at the distance x:
E= Q4π ∈0 x2
Question 13:
Consider the following very rough model of a beryllium atom. The nucleus has four protons and four neutrons confined to a small volume of radius 10^{−15} m. The two 1 s electrons make a spherical charge cloud at an average distance of 1⋅3 ×10^{−11} m from the nucleus, whereas the two 2 s electrons make another spherical cloud at an average distance of 5⋅2 × 10^{−11} m from the nucleus. Find three electric fields at (a) a point just inside the 1 s cloud and (b) a point just inside the 2 s cloud.
Answer:
(a)
Let us consider the three surfaces as three concentric spheres A, B and C.
Let us take q =
1.6×1019 C.
Sphere A is the nucleus; so, the charge on sphere A,
q1 = 4qSphere B is the sphere enclosing the nucleus and the 2 1s electrons; so charge on this sphere,
q2 = 4q 2q = 2qSphere C is the sphere enclosing the nucleus and the 4 electrons of Be; so, the charge enclosed by this sphere,
q3 = 4q4q = 0Radius of sphere A,
r1 = 1015 mRadius of sphere B,
r2 = 1.3×1011mRadius of sphere C,
r3 = 5.2×1011 mAs the point ‘P’ is just inside the spherical cloud 1s, its distance from the centre
x=1.3×1011 mElectric field,
E=q4π ∈0 x2Here, the charge enclosed is due to the charge of the 4 protons inside the nucleus. So,
E=4×1.6×10194×3.14×8.85×1012×1.3×10112E=3.4×1013 N/C(b)
For a point just inside the 2s cloud, the total charge enclosed will be due to the 4 protons and 2 electrons. Charge enclosed,
qen=2q =2×1.6×1019 CHence, electric field,
E=qen4π ∈0 x2x = 5⋅2 × 10^{−11} m
E=2×1.6×10194×3.14×8.85×1012×5.2×10112E=1.065×1012 N/CThus,
E=1.1×1012 N/C
Question 14:
Find the magnitude of the electric field at a point 4 cm away from a line charge of density
2×106 C m1.
Answer:
Given:
Charge density of the line containing charge, λ =
2×106 C m1We need to find the electric field at a distance of 4 cm away from the line charge.
We take a Gaussian surface around the line charge of cylindrical shape of radius r = 4 ×10
2 m and height l.
The charge enclosed by the Gaussian surface, q_{en} = λl
Let the magnitude of the electric field at a distance of 4 cm away from the line charge be E.
Thus, net flux through the Gaussian surface,
ϕ=∮E→.ds→There will be no flux through the circular bases of the cylinder; there will be flux only from the curved surface.
The electric field lines are directed radially outward, from the line charge to the cylinder.
Therefore, the lines will be perpendicular to the curved surface.
Thus,
ϕ=∮E→.ds→=E∮ds⇒ϕ=E×2πrlApplying Gauss’s theorem,
ϕ=qen∈0E × 2πrl=λ l∈0 E=λ2πr ∈0
E=2×1062×3.14×8.85×1012×4×102E=8.99×105 N/C
Question 15:
A long cylindrical wire carries a positive charge of linear density
2·0×108 C m1. An electron revolves around it in a circular path under the influence of the attractive electrostatic force. Find the kinetic energy of the electron. Note that it is independent of the radius.
Answer:
Let the linear charge density of the wire be
λ.
The electric field due to a charge distributed on a wire at a perpendicular distance r from the wire,
E=λ2π∈0 rThe electrostatic force on the electron will provide the electron the necessary centripetal force required by it to move in a circular orbit. Thus,
qE = mv2r⇒mv2 = qEr …(1)Kinetic energy of the electron,
K= 12mv2From (1),
K =qEr2K=qr2λ2π∈0r ∵ E=λ2π∈0r K=(1.6×1019)×(2×108)×(9×109) JK=2.88×1017 J
Question 16:
A long cylindrical volume contains a uniformly distributed charge of density
ρ. Find the electric field at a point P inside the cylindrical volume at a distance x from its axis (figure 30E5).
Figure
Answer:
Given:
Volume charge density inside the cylinder =
ρLet the radius of the cylinder be r.
Let charge enclosed by the given cylinder be Q
Consider a Gaussian cylindrical surface of radius x and height h.
Let charge enclosed by the cylinder of radius x be
q’.
The charge on this imaginary cylinder can be found by taking the product of the volume charge density of the cylinder and the volume of the imaginary cylinder. Thus,
q’=
ρπx2hFrom Gauss’s Law,
∮E.ds=qen∈0E.2πxh=ρ(πx2 h)∈0E=ρx2 ∈0
Question 17:
A nonconducting sheet of large surface area and thickness d contains a uniform charge distribution of density
ρ. Find the electric field at a point P inside the plate, at a distance x from the central plane. Draw a qualitative graph of E against x for 0 < x < d.
Answer:
Given:
Thickness of the sheet = d
Let the surface area of the sheet be s.
Volume of the sheet = sd
Volume charge density of the sheet,
ρ =QsdCharge on the sheet = Q
Consider an imaginary plane at a distance x from the central plane of surface area s.
Charge enclosed by this sheet, q =
ρsxFor this Guassian surface, using Gauss’s Law,we get:
∮ E.ds=q∈0E. s=ρsx∈0 E=ρx∈0The electric field outside the sheet will be constant and will be:
E=ρd∈0
Question 18:
A charged particle with a charge of −2⋅0 × 10^{−6} C is placed close to a nonconducting plate with a surface charge density of
4·0×106 C m2. Find the force of attraction between the particle and the plate.
Answer:
The electric field due to a conducting thin sheet,
E=σ2 ∈0The magnitude of attractive force between the particle and the plate,
F=qEF=q×σ2∈0F=2.0×106×4.0×1062×8.85×1012F=0.45 N
Question 19:
One end of a 10 cm long silk thread is fixed to a large vertical surface of a charged nonconducting plate and the other end is fastened to a small ball of mass 10 g and a charge of
4·0×106 C. In equilibrium, the thread makes an angle of 60° with the vertical. Find the surface charge density on the plate.
Answer:
There are two forces acting on the ball. These are
(1) Weight of the ball, W = mg
(2) Coulomb force acting on the charged ball due to the electric field of the plate, F = qE
Due to these forces,a tension develops in the thread.
Let the surface charge density on the plate be σ.
Electric field of a plate,
E =
σ2∈0It is given that in equilibrium, the thread makes an angle of 60° with the vertical.
Resolving the tension in the string along horizontal and vertical directions, we get:
T cos 60°=mgT sin 60°= qE⇒tan 60°=qEmg⇒ E=mg tan 60°qAlso, electric field due to a plate,
E=σ2 ∈0=mg tan60oqσ=2 ∈0 mg tan 60°q σ=2×8.85×1012×10×103×9.8×1.7324.0×106 σ=7.5×107 C/m2
Question 20:
Consider the situation of the previous problem. (a) Find the tension in the string in equilibrium. (b) Suppose the ball is slightly pushed aside and released. Find the time period of the small oscillations.
Answer:
(a)
In equilibrium state, the thread makes an angle of 60^{o} with the vertical.
The tension in the thread is resolved into horizontal and vertical components.
Then, tension in the string in equilibrium,
T cos 60°=mgT×12=10×103×10T=10×103×10×2=0.20 N(b) As it is displaced from equilibrium, net force on the ball,
F=mg2+qσ2∈02As F = ma
⇒a=g2+qσm2∈02The surface charge density of the plate (as calculated in the previous question), σ = 7.5×10
7 C/m^{2}
Charge on the ball, q = 4×10
6 C
Mass of the ball, m =
The time period of oscillation of the given simple pendulum,
T=2π lg=2π 10×1029.8=0.45 sec
Question 21:
Two large conducting plates are placed parallel to each other with a separation of 2⋅00 cm between them. An electron starting from rest near one of the plates reaches the other plate in 2⋅00 microseconds. Find the surface charge density on the inner surfaces.
Answer:
Distance travelled by the electron, d= 2 cm
Time taken to cross the region, t = 2×10
6 s
Let the surface charge density at the conducting plates be σ.
Let the acceleration of the electron be a.
Applying the 2nd equation of motion, we get:
d=12 at2 ⇒a=2dt2This acceleration is provided by the Coulombic force. So,
a=qEm=2dt2⇒E=2 mdqt2 E=2×(9.1×1031)×(2×102)(1.6×1019)×(4×1012) E=5.6875×102 N/CAlso, we know that electric field due to a plate,
E=σ∈0⇒σ = ∈0E⇒σ=8.85×1012×5.68×102 C/m2⇒σ=50.33×1014 C/m2=0.503×1012 C/m2
Question 22:
Two large conducting plates are placed parallel to each other and they carry equal and opposite charges with surface density
σas shown in the figure (30E6). Find the electric field (a) at the left of the plates (b) in between the plates and (c) at the right of the plates.
Figure
Answer:
Given :
Surface charge density on the plates
=σThe electric field due to plate 1, E_{1} =
σ2∈0The electric field due to plate 2, E_{2} =
σ2∈0(a) The strength of the electric field due to both the plates will be same but their directions will be opposite to each other on any point at the left of the two plates.
Thus, the net electric field at a point on the left of plate1 =
σ2∈0σ2∈0=0(b) Here the direction of the fields will be same. So, they will add up to give the resultant field in this region.
Total electric field:
σ2 ∈0+σ2 ∈0=σ∈0 (c) The strength of the electric field due to both the plates will be same but their directions will be opposite to each other at any point on the right of the two plates.
Thus, the net electric field at a point on the left of plate 2 =
σ2∈0σ2∈0=0
Question 23:
Two conducting plates X and Y, each with a large surface area A (on one side), are placed parallel to each other, as shown in the figure (30E7). Plate X is given a charge Q, whereas the other is kept neutral. Find (a) the surface charge density at the inner surface of plate X (b) the electric field at a point to the left of the plates (c) the electric field at a point in between the plates and (d) the electric field at a point to the right of the plates.
Figure
Answer:
(a)
Given that the charge present on the plate is Q. The other plate will get the same charge Q due to convection.
Let the surface charge densities on both sides of the plate be σ_{1} and σ_{2}.
Now, electric field due to a plate,
E =σ2∈0So, the magnitudes of the electric fields due to this plate on each side
=σ12 ∈0 and σ22 ∈0The plate has two sides, each of area A. So, the net charge given to the plate will be equally distributed on both the sides.This implies that the charge developed on each side will be
q1=q2=Q2This implies that the net surface charge density on each side
=Q2A(b)
Electric field to the left of the plates
On the left side of the plate surface, charge density,
σ=Q2AHence, electric field
=Q2A ∈0This must be directed towards the left, as ‘X’ is the positivelycharged plate.
(c) Here, the charged plate ‘X’ acts as the only source of electric field, with positive in the inner side. Plate Y is neutral. So, a negative charge will be induced on its inner side. ‘Y’ attracts the charged particle towards itself. So, the middle portion E is towards the right and is equal to
Q2A∈0.
(d) Similarly for the extreme right, the outer side of plate ‘Y’ acts as positive and hence it repels to the right with
E=Q2A ∈0
Question 24:
Three identical metal plates with large surface areas are kept parallel to each other as shown in the figure (30E8). The leftmost plate is given a charge Q, the rightmost a charge −2Q and the middle one is kept neutral. Find the charge appearing on the outer surface of the rightmost plate.
Figure
Answer:
Consider the Gaussian surface as shown in the figure.
Let the charge on the outer surface of the leftmost plate be q. Thus, the charges on the plates are distributed as shown in the diagram.
The net field at point P due to all the induced charges must be zero, as it is lying inside the metal surface.
Let the surface area of the plates be A.
Electric field at point P due to the charges on plate X:
Due to charge (+Q
–q) is
Qq2A∈0in the right direction
Due to charge (+q) is
q2A∈0in the right direction
Electric field at point P due to charges on plate Y:
Due to charge (
–q) is
q2A∈0in the right direction
Due to charge (+q) is
q2A∈0in the left direction
Electric field at point P due to charges on plate Z:
Due to charge (
–q) is
q2A∈0in the right direction
Due to charge (
2Q + q) is
2Qq2A∈0in the right direction
The net electric field at point P:
Qq2A∈0+
q2A∈0
– q2A∈0 – q2A∈0+
q2A∈0+
2Qq2A∈0= 0
Qq2A∈0+
2Qq2A∈0= 0
Qq+2Qq=03Qq=0q=3Q2Thus, the charge on the outer plate of the rightmost plate =
2Q+q=2Q+3Q2=Q2
HC Verma Solutions for Class 11 Physics – Part 1
 Chapter 1 – Introduction to Physics
 Chapter 2 – Physics and Mathematics
 Chapter 3 – Rest and Motion: Kinematics
 Chapter 4 – The Forces
 Chapter 5 – Newton’s Laws of Motion
 Chapter 6 – Friction
 Chapter 7 – Circular Motion
 Chapter 8 – Work and Energy
 Chapter 9 – Center of Mass, Linear Momentum, Collision
 Chapter 10 – Rotational Mechanics
 Chapter 11 – Gravitation
 Chapter 12 – Simple Harmonic Motion
 Chapter 13 – Fluid Mechanics
 Chapter 14 – Some Mechanical Properties of Matter
 Chapter 15 – Wave Motion and Wave on a String
 Chapter 16 – Sound Wave
 Chapter 17 – Light Waves
 Chapter 18 – Geometrical Optics
 Chapter 19 – Optical Instruments
 Chapter 20 – Dispersion and Spectra
 Chapter 21 – Speed of Light
 Chapter 22 – Photometry
HC Verma Solutions for Class 12 Physics – Part 2
 Chapter 23 – Heat and Temperature
 Chapter 24 – Kinetic Theory of Gases
 Chapter 25 – Calorimetry
 Chapter 26 – Laws of Thermodynamics
 Chapter 27 – Specific Heat Capacities of Gases
 Chapter 28 – Heat Transfer
 Chapter 29 – Electric Field and Potential
 Chapter 30 – Gauss’s Law
 Chapter 31 – Capacitors
 Chapter 32 – Electric Current in Conductors
 Chapter 33 – Thermal and Chemical Effects of Electric Current
 Chapter 34 – Magnetic Field
 Chapter 35 – Magnetic Field due to a Current
 Chapter 36 – Permanent Magnets
 Chapter 37 – Magnetic Properties of Matter
 Chapter 38 – Electromagnetic Induction
 Chapter 39 – Alternating Current
 Chapter 40 – Electromagnetic Waves
 Chapter 41 – Electric Current through Gases
 Chapter 42 – Photoelectric Effect and Wave Particle Duality
 Chapter 43 – Bohr’s Model and Physics of the Atom
 Chapter 44 – Xrays
 Chapter 45 – Semiconductors and Semiconductor Devices
 Chapter 46 – The Nucleus
 Chapter 47 – The Special Theory of Relativity