• Electric field strength for a radial field - class-XII
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Electric field strength for a radial field - class-XII

Description: electric field strength for a radial field
Number of Questions: 14
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Tags: physics coulomb's law
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A charge is kept at the centre of a shell. Shell has charge Q uniformally distibuted over its surface and radius R. The force on the central charge due to the shell is :

applications of gauss's law coulomb's law physics

  1. towards left

  2. towards right

  3. upward

  4. zero

Show answer
Correct Option: D
Explanation:

Electric field at the centre of shell is zero when charge is uniformally distributed over surface of shell . 

Hence the force on the charge at the centre is zero.

A hollow conducting sphere of charge does not have electric field at

applications of gauss's law coulomb's law physics

  1. outer point

  2. interior point

  3. beyond $2m$

  4. beyond $100m$

Show answer
Correct Option: B
Explanation:

$E=0$, at any point inside the sphere

Assertion: Electric field inside a current carrying wire is zero.
Reason: Net charge in a current carrying wire is non zero.

applications of gauss's law coulomb's law physics

  1. A

  2. B

  3. C

  4. D

Show answer
Correct Option: C

The rupture of air medium occurs at $E=3\times 10^6 \ V/m$. The maximum charge that can be given to a sphere of diameter $5 \ m$ will be (in coulomb):

applications of gauss's law coulomb's law physics

  1. $2\times 10^{-2}$

  2. $2\times 10^{-3}$

  3. $2\times 10^{-4}$

  4. $2\times 10^{-5}$

Show answer
Correct Option: B
Explanation:

Given, $E _o= 3 \times 10^6 \ V/m;$  Diameter of sphere $= 5 \ m$

Electric field on the surface of the sphere $= \cfrac {KQ}{R^2}$

$E _o= \cfrac {KQ}{R^2}$

$Q= \cfrac {R^2E^o}{K}= \cfrac {6.25 \times 3 \times 10^6}{9 \times 10^9}$

$= 2 \times 10^{-3}\ C$

A spherical shell of mass $m$ and radius $R$ filled completely with a liquid of same mass and set to rotate about a vertical axis through its centre has a moment of inertia ${I _1}$ about the axis$.$ The liquid starts leaking out of the hole at the buttom$.$ If moment of inertia of the system is ${I _2}$ when the shell is half filled and ${I _3}$ is the moment of inertia when entire water drained off$.$ then $:$

applications of gauss's law coulomb's law physics

  1. $\dfrac{{{I _1}}}{{{I _2}}} \approx 1.5$

  2. $\dfrac{{{I _{ _1}}}}{{{I _2}}} \approx 0.67$

  3. $\dfrac{{{I _1}}}{{{I _3}}} \approx 1.6$

  4. $\dfrac{{{I _2}}}{{{I _3}}} \approx 1.4$

Show answer
Correct Option: C
Explanation:

Moment of inertia of a shell$,$ ${I _3} = \frac{2}{3}M{R^2}$ 

Moment of inertia of a completely filled sphere$,$ ${I _1} = \frac{2}{3}M{R^2} + \frac{2}{5}M{R^2} = \frac{{16}}{{15}}M{R^2}$ 
Moment of inertia of a half sphere$,$ ${I _2} = \frac{2}{3}M{R^2} + \frac{2}{5}\frac{M}{2}{R^2} = \frac{{26}}{{30}}M{R^2}$
Hence,
option $(C)$is correct answer.

A positive charge q is placed in a spherical cavity made in a positively charged sphere. The centres of sphere cavity are displaced by a small distance $\overrightarrow l $. Force on charge q is:

applications of gauss's law coulomb's law physics

  1. in the direction parallel to vector $\overrightarrow l $

  2. in radial direction

  3. in a direction which depends on the magnitude of charge density in sphere

  4. direction can not be determined.

Show answer
Correct Option: D

At all points inside a uniform spherical shell -

applications of gauss's law coulomb's law physics

  1. gravitational intensity and gravitational potential both are zero

  2. gravitational intensity and gravitational potential both are non- zero

  3. gravitational intensity is non- zero and gravitational potential both are zero

  4. gravitational intensity is zero and gravitational potential both are non-zero

Show answer
Correct Option: A
Explanation:

Gravity Force Inside a Spherical Shell. The net gravitational force on a point mass inside a spherical shell of mass is identically zero! Physically, this is a very important result because any spherically symmetric mass distribution outside the position of the test mass m can be build up as a series of such shells


A thinwalled, spherical conducting shell S of radius R is given charge Q. The same amountof charge is also placed at its centre C. Which of the following statements are correct?

applications of gauss's law coulomb's law physics

  1. On the outer surface of S, the charge density is $\displaystyle \frac {Q} {2 \pi R^2} $

  2. The electric field is zero at all points inside S

  3. At a point just outside S, the electric field is double the field at a point just inside S

  4. At any point inside S, the electric field is inversely proportional to the square of its distance from C

Show answer
Correct Option: A,C,D
Explanation:

because of charge Q at center there will be induced charge -Q at inner surface of sphere. hence charge density $\dfrac{2Q}{4\pi r^2} = \dfrac{Q}{2\pi r^2}$

because of 2Q charge outside the electric field is double that of inside.
At any point inside S, the electric field is inversely proportional to the square of its distance from C

Two sphere's are isolated from each other. They each have an identical net positive charge and have the same radius, however, one sphere is solid and insulating, while the other is a hollow conducting sphere whose charge is uniformly distributed.
For which sphere is the electric field the greatest distance $x$ from the center of the spheres?
Assume $x$ is less than the radius of the spheres.

applications of gauss's law coulomb's law physics

  1. The conducting hollow sphere has a greater E-field.

  2. The insulating solid sphere has a greater E field.

  3. Both spheres have the same E field.

  4. Neither sphere would cause there to be an Electric field.

Show answer
Correct Option: B
Explanation:

The charge in a conducting hollow sphere resides on the surface only. Hence on applying Gauss Law on a spherical surface enclosed in the sphere, we get electric field to be zero inside it.

For a solid insulating sphere with uniform charge distribution, finite electric field exists in the sphere.
Hence correct answer is option B.

As one penetrates through uniformly charged conducting sphere, what happens to the electric field strength:

applications of gauss's law coulomb's law physics

  1. decreases inversely as the square of the distance

  2. decreases inversely as the distance

  3. becomes zero

  4. increases inversely as the square of distance

Show answer
Correct Option: C
Explanation:

Electric field strength inside the uniform charged sphere is zero.

$\therefore$  As one penetrates through uniformly charged sphere, electric field strength inside the sphere becomes zero. 

The magnitude of the electric field on the surface of a sphere of radius $r$ having a uniform surface charge density $\sigma$ is

applications of gauss's law coulomb's law physics

  1. $\sigma / \epsilon _{0}$

  2. $\sigma / 2\epsilon _{0}$

  3. $\sigma / \epsilon _{0}r$

  4. $\sigma / 2\epsilon _{0}r$

Show answer
Correct Option: A
Explanation:
The magnitude of the electric field on the surface of radius $=r$
Charge density $=6$
Then, $E=\dfrac { 6 }{ { \epsilon  } _{ 0 } } $
The electric field is independent of the surface radius.

Consider a thin spherical shell of radius $R$ consisting of uniform surface charge density $\sigma$. The electric field at a point of distance $x$ from its centre and outside the shell is

applications of gauss's law coulomb's law physics

  1. inversely proportional to $\sigma$

  2. directly proportional to ${x}^{2}$

  3. directly proportional to $R$

  4. inversely proportional to ${x}^{2}$

Show answer
Correct Option: D
Explanation:
For a thin uniformly charged spherical shell, the field points outside the shell at a distance $x$ from the centre is
$E=\cfrac { 1 }{ 4\pi { \varepsilon  } _{ 0 } } \cfrac { Q }{ { x }^{ 2 } } $
If the radius of the sphere is $R,Q=\sigma 4\pi { R }^{ 2 }$
$\therefore E=\cfrac { 1 }{ 4\pi { \varepsilon  } _{ 0 } } \cfrac { \sigma 4\pi { R }^{ 2 } }{ { x }^{ 2 } } =\cfrac { \sigma { R }^{ 2 } }{ { { \varepsilon  } _{ 0 }x }^{ 2 } } $
This is inversely proportional to square of the distance from the centre. It is as if the whole charge is concentrated at the centre

Two charged spheres having radii a and b are joined with a wire then the ratio of electric field $\dfrac{E _a}{E _b}$ on their surface is?

applications of gauss's law coulomb's law physics

  1. a/b

  2. b/a

  3. ba

  4. None of these

Show answer
Correct Option: B
Explanation:

When the two spheres are connected by a wire, then both of them acquire the same potential say $V$.


We also know that the electric field on the surface of a sphere $E=\dfrac{Q}{4\pi\epsilon _o r^2}$
and potential on the surface is given by $V=\dfrac{Q}{4\pi\epsilon _or}$

$\implies E=\dfrac{V}{r}$

Here, V is constant , hence  $E\propto \dfrac{1}{r}$

$\implies \dfrac{E _a}{E _b}=\dfrac{b}{a}$

Charges $Q _1$ and $Q _2$ are placed inside and outside respectively of an uncharged conducting shell. Their seperation is r.

applications of gauss's law coulomb's law physics

  1. The force on $Q _1$ is zero.

  2. The force on $Q _1$ is $\displaystyle k \frac{Q _1 Q _2}{r^2}$

  3. The force on $Q _2$ is $\displaystyle k \frac{Q _1 Q _2}{r^2}$

  4. The force on $Q _2$ is zero.

Show answer
Correct Option: A,C
Explanation:

As the electric field inside the conducting shell is zero , so the force on the inner charge, $Q _1$ will be zero.
The electric field at outside charge $Q _2$ due to $Q _1$ is $E=k\frac{Q _1}{r^2}$
Force on $Q _2$ is $F=Q _2E=k\frac{Q _1Q _2}{r^2}$

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