Tag: coulomb's law

Questions Related to coulomb's law

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

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

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

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

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


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

  1. inversely proportional to $\sigma$

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

  3. directly proportional to $R$

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


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?

  1. a/b

  2. b/a

  3. ba

  4. None of these


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.

  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.


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 $F=Q _2E=k\frac{Q _1Q _2}{r^2}$

If we move in a direction opposite to the electric lines of force:

  1. electrical potential decreases

  2. electrical potential increases

  3. electrical potential remains uncharged

  4. nothing can be said.


Correct Option: B

A uniform wire $10 \,cm$ long is carrying a steady current. The potential drop across it is $10V$. The electric field inside it is _____

  1. zero

  2. $1Nm^{-1}$

  3. $10 \,Vm^{-1}$

  4. $100 \,Vm^{-1}$


Correct Option: D
Explanation:

$E = \dfrac{V}{r} = \dfrac{10V}{10 \times 10^{-2}m} = \dfrac{10 \times 100}{10} = 100 \dfrac{v}{m}$

The electric potential while moving along the lines of force

  1. decreases

  2. increases

  3. remains constant

  4. becomes infinite


Correct Option: A

$E=-\dfrac{dV}{dr}$, here negative sign signified that

  1. E is opposite to V

  2. E is negative

  3. E increases when V decreases

  4. E is directed in the direction of decreasing V


Correct Option: D
Explanation:

The negative sign is just a convention and it signifies that the direction of E is opposite to the direction in which potential increases.

The ratio of electric force $ ( F _e ) $ to gravitational force acting between two electrons will be:

  1. $ 1 \times 10^{36} $

  2. $ 2 \times 10^{39} $

  3. $ 2.5\times 10^{39} $

  4. $ 3 \times 10^{39} $


Correct Option: C

Electric potential at ( x, y, z ) is given as $V$= $- x ^ { 2 } y \sqrt { z }$ Find the electrical field at (2 ,1, 1)

  1. $4 \hat { i } + 4 \hat { j } + 4 \hat { k }$

  2. $- 4 \hat { i } - 4 \hat { j } - 2 \hat { k }$

  3. $- 4 \hat { i } - 4 \hat { j } - 4 \hat { k }$

  4. $4 \hat { 1 } + 4 \hat { j } + 2 \hat { k }$


Correct Option: D