Physics

Electrostatics

303 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

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

The electric field of a uniformly charged disc along its axis is directed away from the disc. A positive charge q approaching the disc will experience a repulsive force, causing it to decelerate and potentially return before reaching the disc.

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

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. $2.83$ $J/m^3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Energy density = $=(\frac { 1 }{ 2 } )∈0E2$ 
$=(\frac { 1 }{ 2 } )\times 8.86\times 10-12\times \left( \frac { v }{ r }  \right) 2$
$=4.43\times 10-12\times [(8000)/(10-2)]2$
$=283.5\times 10-2$
$=2.83J/m3$
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 } $      ($\because$   $\dfrac { 1 }{ \pi { \epsilon  } _{ 0 } } =K$)
Multiple choice potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

A uniform electric field of magnitude $290 V/m$ is directed in the positive $x$ direction. A $+13.0 \mu C$ charge moves from the origin to the point $(x, y) = (20.0 cm, 50.0 cm).$

What is the change in the potential energy of the charge field system?

  1. $-754J$
  2. $-754mJ$
  3. $-754kJ$
  4. $-754\mu J$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given electric field, $\overrightarrow{E}=290\, V/m$ directed along $+x$ direction.
Charge $=+13\mu C$ moves from origin to point $(x,y)=(20\, cm , 50\, cm)$.
We have to find the charge in potential energy of the charge field system.
We know, the change in potential energy  $=$ charge $\times $ change in potential
i.e, $\Delta U=q\Delta V$
But, $\Delta V=-Ed$
$\Delta U=-qEd$
$=-13\times 10^{-6}\times 290\times 20\times 10^{-2}$
$=0.000754\, J$
$=-754\times 10^{-6}\, J$
Here $d=20\, cm$, since electric field is along positive $x-$axis.
Multiple choice physics static electricity properties of charges charge properties of charge

Choose best option about cavity for a charged metallic sphere: 

  1. Net field inside the cavity is zero

  2. Net field inside the metal is non-zero constant

  3. Potential inside the metal is constant

  4. Potential outside metal is constant
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Inside the cavity of charged measure sphere net electric field is non zero constant.

$\therefore $ Option $A$ is correct.

Multiple choice physics static electricity properties of charges charge properties of charge

Two charges 2$\mathrm { uC }$ and 1$\mathrm { \mu }C$ are placed at adistance of 10$\mathrm { cm }$ . The position of the third charge from2$\mu \mathrm { C }$ between them so that it does not experience any force

  1. $7cm$
  2. $2cm$
  3. $5.858cm$
  4. $8cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a charge to experience no force, the electric fields from the two charges must be equal and opposite. Setting k(2)/x^2 = k(1)/(10-x)^2 leads to 2(10-x)^2 = x^2, which simplifies to sqrt(2)(10-x) = x. Solving for x gives x = 10*sqrt(2)/(1+sqrt(2)) approximately 5.858 cm.

Multiple choice physics static electricity properties of charges charge properties of charge

An infinite number of charges each equal to q coulomb are placed along x-axis at x 1, x = 2, x= 8. so on the potential and electric field at x =0 due to this arrangement is

  1. $\dfrac { q } { 2 \pi \varepsilon _ { 0 } } , \dfrac { 3 q } { 4 \pi \varepsilon _ { 0 } }$
  2. $\dfrac { q } { 2 \pi \varepsilon _ { 0 } } \cdot \frac { q } { 3 \pi \varepsilon _ { 0 } }$
  3. $\dfrac { 2 q } { \pi \varepsilon _ { 0 } } , \dfrac { q } { 3 \pi \varepsilon _ { 0 } }$
  4. $\dfrac { q \varepsilon _ { 0 } } { \pi 2 } , \dfrac { q E _ { 0 } } { 3 \pi }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice physics static electricity properties of charges charge properties of charge

What is the minimum possible amount of charge?

  1. Electronic charge $e$
  2. Electronic charge $2e$
  3. Electronic charge $\dfrac{e}{2}$
  4. Electronic charge $\dfrac{e}{\sqrt{2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The minimum possible charge is equal to the charge of the electron.That is, $e=1.6\times { 10 }^{ -19 }C$.
The electron is a subatomic particle, symbol e, with a negative elementary electric charge. Electrons belong to the first generation of the lepton particle family, and are generally thought to be elementary particles because they have no known components or substructure.

Multiple choice physics static electricity properties of charges charge properties of charge

The charge on a body is +1 C. Find the number of electrons in excess or deficit on the body.

  1. 6.25 $\times 10^{1}$ coulomb
  2. 6.25 $\times 10^{-18}$ coulomb
  3. 6.25 $\times 10^{-1}$ coulomb
  4. 6.25 $\times 10^{18}$ coulomb
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The SI unit of electric charge is Coulomb, One coulomb is equal to about $6.242\times { 10 }^{ 18 }e$. 
One coulomb is defined as the quantity of charge which will pass through the cross section of an electrical conductor while one ampere current is flowing through the conductor in one second. Electric charge is denoted by Q.