Physics

Electrostatics

299 Questions

Electrostatics deals with electric charges, fields, and potentials at rest. It is a crucial topic for physics sections in engineering and civil services competitive examinations. Review these questions to build a strong understanding of Coulomb law, electric dipoles, Gauss law, and electric flux.

Electric field and potentialElectric dipole momentGauss Law applicationsCoulomb force calculationsCharge distribution on spheresEquipotential surfaces

Electrostatics Questions

Multiple choice potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

what is the potential difference between two points, if 2J of work must be done to move a 4 mC charge from one point to another is:

  1. 50 V

  2. 500 V

  3. 5 V

  4. 5000 V
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Answer is B.

The total work done = energy transferred.
so, we might see the equation energy = voltage x charge, E = V * Q, written as, 
work = voltage x charge, W = V * Q.
In this case, the charge is 4 mC, that is, 0.004 C and work done is 2 J.
Therefore, V=W/Q = 2/0.004 = 500 V.
Hence, the potential difference between two points if 2 J of work must be done to move a 4 mC charge from one point to another is 500 V.

Multiple choice potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

A point charge q is rotated along a circle in the electric field generated by another point charge Q. The work done by the electric field on the rotating charge in one complete revolution is_______

  1. zero

  2. positive

  3. negative

  4. zero if the charge Q is at the center and nonzero otherwise.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The net displacement round one complete circle is 0.

So, the work done is 0.

Multiple choice potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

$100J$ of work is done when $2 \mu C$ charge is moved in an electric field between two points. The p.d. between the points is

  1. $2\times10^{-4}V$
  2. $2\times10^{-8}V$
  3. $2\times10^{-6}V$
  4. $5\times10^{7}V$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know work done $=q\Delta V $
Where $\Delta V $ is change in potential $V _2-V _1$
$100=\Delta V\times 2\times 10^{-6}$
$50\times 10^6=\Delta V$
$\therefore p\cdot d= 5\times 10^7 V$

Multiple choice potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

A hollow metal sphere of radius $5\ cm$ is charged such that the potential on its surface is $10$ volts. The potential of the centre of the sphere is:

  1. zero
  2. $10$ volts
  3. Same as at a point $5\ cm$ away from the surface.
  4. Same as at a point $25\ cm$ away from the surface.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
For hollow spherical conductor -
$\Rightarrow$ Potential on its surface $=$ Potential on center $= 10\;volts.$
Because, the electric field inside the spherical shell is zero, due to which the potential inside the shell is constant and will be equal to the potential at the surface.
Hence, the answer is $10\;volts.$
Multiple choice potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

A spherical shell of radius $R _1$ with uniform charge $q$ is expanded to a radius $R _2$. Find the work performed by the electric forces during the shell expansion from $R _1$ to radius $R _2$.

  1. $\dfrac{q^2}{2\pi in _0}\left(\dfrac{1}{R _1} - \dfrac{1}{R _2}\right)$
  2. $\dfrac{q^2}{3\pi in _0}\left(\dfrac{1}{R _1} - \dfrac{1}{R _2}\right)$
  3. $\dfrac{q^2}{5\pi in _0}\left(\dfrac{1}{R _1} - \dfrac{1}{R _2}\right)$
  4. $\dfrac{q^2}{8\pi in _0}\left(\dfrac{1}{R _1} - \dfrac{1}{R _2}\right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Spherical shell of radius $={ R } _{ 1 }$
charge $=q$
work done $=W=q\left( { V } _{ B }-{ V } _{ A } \right) $
${ V } _{ A }=\dfrac { { K } _{ q } }{ { R } _{ 1 } } $
${ V } _{ B }=\dfrac { { K } _{ q } }{ { R } _{ 2 } } $
$W=q\left( \dfrac { { K } _{ q } }{ { R } _{ 1 } } -\dfrac { { K } _{ q } }{ { R } _{ 2 } }  \right) $
     $=\dfrac { { q }^{ 2 } }{ 8\pi { \epsilon  } _{ 0 }{ n } _{ 0 } } \left( \dfrac { 1 }{ { R } _{ 1 } } -\dfrac { 1 }{ { R } _{ 2 } }  \right) $    (Proved).
Multiple choice potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

Two insulated charged spheres of radii ${R} _{1}$ and ${R} _{2}$ having charges ${Q} _{1}$ and ${Q} _{2}$ respectively are connected to each other, then there is:

  1. no change in the energy of the sytem

  2. an increase in the energy of the system

  3. always a decrease in the energy of the system
  4. a decrease in energy of the system unless ${q} _{1}{R} _{2}={q} _{2}{R} _{1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Two insulated charged spheres of radii $ R _1,R _2$having charges $q _1,q _2$respectively are brought in contact with each other.

Charges will flow across the point of contact from where they are connected until their potential at their surfaces became same. 
Energy is decreased because system changes to get stability.
Potential at their surfaces are same,
So,
$V _1=V _2$
$\dfrac{Kq _1}{R _1}=\dfrac{Kq _2}{R _2}$
$q _1R _2=q _2R _1$
Hence Proved.

Multiple choice potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

Calculate the electrostatic potential energy of two electrons separated bt 3 $\overset { \circ  }{ A } $ in vacuum

  1. $7.69\times10^{-19} J$
  2. $8.69\times10^{-19} J$
  3. $5.69\times10^{-19} J$
  4. $6.69\times10^{-19} J$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

U = k * q1 * q2 / r. k = 9 * 10^9, q = 1.6 * 10^-19 C, r = 3 * 10^-10 m. U = (9 * 10^9 * (1.6 * 10^-19)^2) / (3 * 10^-10) = 7.68 * 10^-19 J. Option B is the closest value provided.

Multiple choice potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

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] $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The total potential energy of a cube of 8 charges is the sum of interactions: 12 edges (distance a), 12 face diagonals (distance a*sqrt(2)), and 4 body diagonals (distance a*sqrt(3)). Summing these gives U = (q^2 / (4*pi*e0*a)) * (12 + 12/sqrt(2) + 4/sqrt(3)). Factoring out 12 gives the expression in option D.

Multiple choice potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

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. $-2.35 J$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

U = k * q1 * q2 / r. k = 9 * 10^9, q1 = 10^-5 C, q2 = -10^-5 C, r = 0.4 m. U = (9 * 10^9 * -10^-10) / 0.4 = -0.9 / 0.4 = -2.25 J.

Multiple choice potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The self-potential energy of a sphere with non-uniform charge density is found by integrating the energy density or using the potential at each shell. For rho = rho0 * x, the integration leads to the result in option C.

Multiple choice potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

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. $\dfrac{1}{4 \pi {\epsilon} _{0}}$ . $\dfrac{2q}{d}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The potential at a point midway between two equal and opposite charges is the sum of the potentials from each: V = k * q / d + k * (-q) / d = 0.

Multiple choice potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

The potential in certain region is given as $V = 2x^2$, then the charge density of that region is 

  1. $-\dfrac{4x}{\varepsilon _0}$
  2. $-\dfrac{4}{\varepsilon _0}$
  3. $-4 \varepsilon _0$
  4. $-2 \varepsilon _0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

By Gauss's law, $\nabla^2V=-\frac{\rho}{\varepsilon _0}$

So, $\frac{\partial^2V}{\partial x^2}=-\frac{\rho}{\varepsilon _0} ...(1)$
Give, $V=2x^2$
or $\frac{\partial V}{\partial x}=4x$
or $\frac{\partial^2V}{\partial x^2}=4$
Now from (1), $\rho=-4\varepsilon _0$

Multiple choice potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

Positive charge Q is uniformly distributed throughout the volume of a dielectric sphere of radius R. A point mass having charge +q and mass m is fired towards the centre of the sphere with velocity v from a point A at distance r(r> R) from the centre of the sphere. Find the minimum velocity v so that it can penetrate R/2 distance of the sphere. Neglect any resistance other than electric interaction. Charge on the small mass remains constant throughout the motion.

  1. $\displaystyle \left[\frac{1}{2 \pi \varepsilon _0} \frac{Qq}{Rm} \left(\frac{r-R}{r} + \frac{3}{4}\right) \right]^{1/2}$
  2. $\displaystyle \left[\frac{1}{2 \pi \varepsilon _0} \frac{Qq}{Rm} \left(\frac{r-R}{r} + \frac{3}{8}\right) \right]^{1/2}$
  3. $\displaystyle \left[\frac{1}{2 \pi \varepsilon _0} \frac{Qq}{Rm} \left(\frac{r-R}{r} - \frac{3}{8}\right) \right]^{1/2}$
  4. $\displaystyle \left[\frac{1}{4 \pi \varepsilon _0} \frac{Qq}{Rm} \left(\frac{r-R}{r} + \frac{3}{4}\right) \right]^{1/2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Initial energy are kinetic and potential energy given by-


$K _i=\dfrac{1}{2}mv^2$       and $U _i=\dfrac{Qq}{4\pi\epsilon _o r}$

Finally at last point , its velocity get reduced to $0$, and potential at a point inside sphere is given by-

$V=\dfrac{Q}{4\pi\epsilon _o}\dfrac{3R^2-r^2}{2R^3}$

At $r=\dfrac{R}{2}$, $U _f=qV$

$\implies U _f=\dfrac{Qq}{4\pi\epsilon _o}\dfrac{3R^2-\dfrac{R^2}{4}}{2R^3}$

$\implies U _f=\dfrac{11}{8}\dfrac{Qq}{4\pi\epsilon _oR}$

Now applying conservation of mechanical energy-

$K _i+U _i=K _f+U _f$

$\implies \dfrac{1}{2}mv^2+\dfrac{Qq}{4\pi\epsilon _o r}=0+\dfrac{11}{8}\dfrac{Qq}{4\pi\epsilon _o R}$

$\implies mv^2=\dfrac{Qq}{2\pi\epsilon _o}\left(\dfrac{11}{8R}-\dfrac{1}{r}\right)$

$\implies v^2=\dfrac{Qq}{2\pi\epsilon _oRm}\left(\dfrac{11}{8}-\dfrac{R}{r}\right)$

$\implies v^2=\dfrac{Qq}{2\pi\epsilon _oRm}\left(1+\dfrac{3}{8}-\dfrac{R}{r}\right)$

$\implies v^2=\dfrac{1}{2\pi\epsilon _o}\dfrac{Qq}{Rm}\left(\dfrac{r-R}{r}+\dfrac{3}{8}\right)$

$\implies v=\sqrt{\dfrac{1}{2\pi\epsilon _o}\dfrac{Qq}{Rm}\left(\dfrac{r-R}{r}+\dfrac{3}{8}\right)}$

Answer-(B)

Multiple choice potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

The electric potential energy of a uniformly charged thin spherical shell of radius 'R' having a total charge 'Q' is

  1. $\dfrac{KQ^2}{4R}$
  2. $\dfrac{KQ^2}{6R}$
  3. $\dfrac{KQ^2}{8R}$
  4. $\dfrac{KQ^2}{16R}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
The electric potential energy of a uniform charged.
radius $=R$
charge $=Q$
electric potential
$E=\dfrac { 1 }{ 2 } \left( \dfrac { 1 }{ 4\pi { \epsilon  } _{ 0 } } .\dfrac { { Q }^{ 2 } }{ R }  \right) $
   $=\dfrac { 1 }{ 8 } K\dfrac { { Q }^{ 2 } }{ R } $      ($\because$   $\dfrac { 1 }{ \pi { \epsilon  } _{ 0 } } =K$)