Tag: electrostatics

Questions Related to electrostatics

A flat circular fixed disc has a charge +Q uniformly distributed on the disc. A charge +q is thrown with kinetic energy K,towards the disc along its axis The charge is q

  1. will not hit the disc at the center

  2. may return back along its path after touching the disc

  3. may return back along its path without touching the disc

  4. any of the above three situation is possible depending on the magnitude of K


Correct Option: A

Particle A having positive charge is moving directly head on towards initially stationary positively charged particle B. At the instant when A and B are closest together. 

  1. the momenta of A and B must be equal

  2. the velocities of A and B must be equal

  3. B would have gained less kinetic energy than A would have lost

  4. B would have gained the same momentum as A would have lost


Correct Option: C,D

Eight charges (each $q$) are placed at the vertices of a regular cube of side $a$. The electric potential energy of the configuration will be $ U=12\times \dfrac { 1 }{ 4\pi \varepsilon _{ 0 } } ,\dfrac { q^{ 2 } }{ a } \times \quad x $ then x.

  1. $ 1+\dfrac { 1 }{ \sqrt { 2 } } +\dfrac { 1 }{ \sqrt { 3 } } $

  2. $ 1+\dfrac { 2 }{ \sqrt { 2 } } +\dfrac { 1 }{ \sqrt { 3 } } $

  3. $ 1+\dfrac { 2 }{ \sqrt { 2 } } +\dfrac { 2 }{ \sqrt { 3 } } $

  4. $ \left[ 1+\dfrac { 1 }{ \sqrt { 2 } } +\dfrac { 1 }{ 3\sqrt { 3 } } \right] $


Correct Option: D

Two point charges of +10 $\mu c$ and -10 $\mu c$ are placed at a distance $40$ cm in air. Potential energy of the system will be-

  1. $2.25 J$

  2. $2.35 J$

  3. $-2.25 J$

  4. $-2.35 J$


Correct Option: C

A solid non-conducting sphere of radius $R$ having charge density $\rho = \rho _{0}x$, where $x$ is distance from the centre of sphere. The self potential energy of the sphere is

  1. $\dfrac {\pi \rho _{0}^{2} R^{4}}{6\epsilon _{0}}$

  2. $\dfrac {\pi \rho _{0}^{2} R^{6}}{4\epsilon _{0}}$

  3. $\dfrac {\pi \rho _{0}^{2} R^{6}}{6\epsilon _{0}}$

  4. None of these


Correct Option: C

If a proton and an electron are accelerated through the same potential difference:

  1. both the proton and electron have same K.E

  2. both the proton and electron have same momentum

  3. both the proton and electron have same velocity

  4. both the proton and electron have same temperature


Correct Option: A
Explanation:

P.D = V
K.E = eV
Velocity will be equal and opposite

What is the change in potential energy of a particle of charge +q that is brought from a distance of 3r to a distance of 2r by a particle of charge q? 

  1. $kq^2/r$

  2. $-kq^2/6r$

  3. $kq^2/r^2$

  4. $-kq^2//4r^2$

  5. $8kq^2/r^2$


Correct Option: B
Explanation:

As the $+q$ is brought near by charge $q$ so there is an attraction force it means q is a negative charge, equal to $-q$.
 The electric potential energy of a system of two point charges (  $+q$ ,  $-q$) sepatrated by a distance $3r$ is given by ,
                  $U=k\dfrac{q\times -q}{3r}$  , 
when the separation is reduced to $2r$ then potential energy will become ,
                 $U'=k\dfrac{q\times- q}{2r}$ 
therefore change in potential energy is 
                             $\Delta U=U'-U$ 
or                         $\Delta U=-k\dfrac{q\times q}{2r}+k\dfrac{q\times q}{3r}$
or                         $\Delta U=kq^{2}\left(1/3r-1/2r\right)=-kq^{2}/6r$

The ratio of momentum of an electron and an alpha particle which are accelerated from rest by  potential difference of 100 V is:

  1. $\sqrt{\dfrac{m _{\alpha}}{m _e}}$

  2. $\sqrt{\dfrac{m _e}{m _{\alpha}}}$

  3. $\dfrac{2m _e}{m _{\alpha}}$

  4. $\sqrt{\dfrac{m _e}{2m _{\alpha}}}$


Correct Option: D
Explanation:

Mass of proton $=1$ $u$

and, Mass of alpha particle $=4$ $u$
Charge of proton $=e$
And, charge of alpha particle $=2e$
For charge moving in a uniform potential field.
$K.E=$ Electric potential energy
$\cfrac{1}{2}mv^2=qV$
$v=\sqrt{\cfrac{2qV}{m}}$
OR,
Momentum $(P)=mv$
$\Rightarrow P=m\sqrt{\cfrac{2qV}{m}}$
$\Rightarrow P=\sqrt{m2qV}$
Momentum of proton,
$P _1=\sqrt{2qVm _1}=\sqrt{2eVm _e}$
Momentum of alpha particle 
$P _2=\sqrt{2qVm _2}=\sqrt{2\times 2eVm _a}$
$\therefore \cfrac{P _1}{P _2}=\sqrt{\cfrac{2eVm _e}{4eVm _a}}$
$\therefore \cfrac{P _1}{P _2}=\sqrt{\cfrac{m _e}{2m _a}}$

A sphere of radius $1$ cm has potential of $8000$V. The energy density near the surface of sphere will be?

  1. $64\times 10^5$ $J/m^3$

  2. $8\times 10^3$ $J/m^3$

  3. $32$ $J/m^3$

  4. $2.83$ $J/m^3$


Correct Option: D
Explanation:
Energy density = $=(\frac { 1 }{ 2 } )∈0E2$ 
$=(\frac { 1 }{ 2 } )\times 8.86\times 10-12\times \left( \frac { v }{ r }  \right) 2$
$=4.43\times 10-12\times [(8000)/(10-2)]2$
$=283.5\times 10-2$
$=2.83J/m3$

Two unlike charges of magnitude q are separated by a distance 2d. The potential at a point midway between them is

  1. zero

  2. $\dfrac{1}{4 \pi {\epsilon} _{0}}$

  3. $\dfrac{1}{4 \pi {\epsilon} _{0}}$ . $\dfrac{q}{d}$

  4. $\dfrac{1}{4 \pi {\epsilon} _{0}}$ . $\dfrac{2q}{d}$


Correct Option: A