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 physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

A point charge q moves from point P to a point S along a path PQRS in a uniform electric field E pointing parallel to the x-axis. The coordinates of P, Q. R and S are $(a, b, 0), (2a, 0, 0), (a, -b, 0)$ and $(0, 0, 0)$. The work done by the field in the above process is :

  1. $zero$
  2. $qEB$
  3. $qEa$
  4. $-qEa$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As the field E is uniform, so E is constant at every point.
As E is directed parallel to x axis, so $\vec{E}=E\hat i$
The work done , $W=\int \vec{F}.\vec{dr}=\int qE\hat i.(\hat{i}dx+\hat{j}dy+\hat{k}dz)$
$W=qE\int dx=qE[\int _a^{2a}dx+\int^a _{2a}dx+\int _a^{0}dx]=qE[2a-a+a-2a+0-a]=-qEa$

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

In a certain region of space, the potential is given by : $V = k {[2x^2 - y^2 + z^2]}$. The electric field at the point (1, 1, 1) has magnitude = 

  1. $k\sqrt{6}$
  2. $2k\sqrt{6}$
  3. $2k\sqrt{3}$
  4. $4k\sqrt{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The electric field is E = -grad(V). E = -(dV/dx i + dV/dy j + dV/dz k). Given V = k(2x^2 - y^2 + z^2), E = -k(4x i - 2y j + 2z k). At (1, 1, 1), E = -k(4 i - 2 j + 2 k). The magnitude is k * sqrt(4^2 + (-2)^2 + 2^2) = k * sqrt(16 + 4 + 4) = k * sqrt(24) = 2k * sqrt(6).

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

A charge of 3C moving in a uniform electric field experiences a force of $3000 N$. The potential difference between two points situated in the field at a distance $1 cm$ from each other will be

  1. $10 V$
  2. $90 V$
  3. $1000 V$
  4. $9000 V$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The electric field E = F/q = 3000 N / 3 C = 1000 N/C. The potential difference V = E * d. With d = 1 cm = 0.01 m, V = 1000 * 0.01 = 10 V.

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

The potential at a point $x$ (measured in $\mu m )$ due to somecharges situated on the $x$ -axis is given by $V ( x ) = 20 / \left( x ^ { 2 } - 4 \right)$Volts. The electric field $E$ at $x = 4 \mu m$ is given by

  1. 5$/ 3$ Volt / \mum and in the -ve $x$ direction
  2. 5$/ 3$ Volt $/ \mu m$ and in the +ve $x$ direction
  3. 10$/ 9$ Volt / \mum and in the -ve $x$ direction
  4. 10$/ 9$ Volt $/ \mu m$ and in the +ve $x$ direction
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The electric field in the x-direction is given by E = -dV/dx. Given V(x) = 20 / (x^2 - 4), differentiating with respect to x using the quotient rule gives dV/dx = 20 * (-1) * (2x) / (x^2 - 4)^2 = -40x / (x^2 - 4)^2. At x = 4, dV/dx = -40(4) / (16 - 4)^2 = -160 / (12^2) = -160 / 144 = -10/9. Since E = -dV/dx, E = -(-10/9) = +10/9 Volt/um, and the positive sign indicates it is in the positive x direction.

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

Variation in potential is maximum if one goes :

  1. along the line of force

  2. perpendicular to the line of force

  3. in any direction

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$dV=-\vec{E}.d\vec{r}=-Edr cos\theta$


Hence, variation will be maximimum for $\theta=0^{o}$ or $180^{o}$, that is variation $dV$ is maximum along line of field or say line of force.

Answer-(A)

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

The electric field lines are closer together near object $A$ than they are near object $B$. We can conclude that :

  1. the potential near $A$ is greater than the potential near $B$
  2. the potential near $A$ is less than the potential near $B$
  3. the potential near $A$ is equal to the potential near $B$
  4. nothing about the relative potentials near $A$ and $B$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Potential decreases in the direction of electric field. So it depends  on whether the lines of forces are from $A$ to $B$ or from $B$ to $A$.

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

There is an electric field $E$ in the x-direction. If the work done by the electric field in moving a charge of $0.2 C$ through a distance of $2 m$ along a line making an angle $60^{\circ}$ with the x-axis is $4 J$, then what is the value of $E$?

  1. $\displaystyle \sqrt3 NC^{-1}$
  2. $\displaystyle 4 NC^{-1}$
  3. $\displaystyle 5 NC^{-1}$
  4. $\displaystyle 20 NC^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\displaystyle F = qE$
work will only be done in moving the charged particle in $x$ direction

work done in moving the charge in y-direction will be $0$
Work done , $W=\int \vec{F}.\vec{dr}$

$\displaystyle W = qE \times 2 cos  60^{\circ}$

or $\displaystyle 4 = 0.2E\times 2 \times \dfrac{1}{2}$

$  \implies  E = 20 NC^{-1}$

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

Charge $Q$ is given a displacement $\displaystyle \vec{r} = a\hat{i}+b\hat{j}$ in an electric field $\displaystyle \vec{E} = E _1\hat{i}+E _2\hat{j}$. The work done is :

  1. $\displaystyle Q(E _1a+E _2b)$
  2. $\displaystyle Q\sqrt{(E _1a)^2+(E _2b)^2}$
  3. $\displaystyle Q (E _1+E _2) \sqrt{a^2+b^2}$
  4. $\displaystyle Q \sqrt{(E _1^2+E^2 _2)^2} \sqrt{a^2+b^2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Work done in the presence of electric field E is $W=\vec F. \vec r = q\vec E.\vec r$
$W=Q[(E _1\hat{i}+E _2\hat{j}).( a\hat{i}+b\hat{j})]$
$W=Q(E _1a+E _2b)$

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

The electric potential decreases uniformly from $120V$ to $80V$ as one moves on the x-axis from $x=-1cm$ to $x=+1cm$. The electric field at the origin

  1. must be equal to $20V{cm}^{-1}$
  2. may be equal to $20V{cm}^{-1}$
  3. may be greater than $20V{cm}^{-1}$
  4. may be less than $20V{cm}^{-1}$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

The electric potential decreases uniformly from 120 V to 80 V as one moves on the x-axis from $x = -1\ cm$ to $ x = +1 \ cm$. The electric field at the origin.

  1. must be equal to 20 Vcm$^{-1}$
  2. must be equal to 20 Vm$^{-1}$
  3. greater than or equal to 20 Vcm$^{-1}$
  4. may be less than 20 Vcm$^{-1}$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

$dv=-\vec E.\vec dx=-E dx \cos\theta$

$\displaystyle E = -\dfrac{dV}{\cos\theta dx} $

$\cos\theta\approx 1$, if we take $\cos\theta =1$,then

$E _{min}= -\dfrac{80-120}{1-(-1)} =\dfrac{40}{2}= 20 V cm^{-1}$

Hence E would be greater than or equal to $20$

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

For a uniform electric field $\vec{E}=E _{0}(\hat{i})$, if the electric potential at x=0 is zero, then the value of electric potential at x=+x will be .......

  1. $xE _{0}$
  2. -$xE _{0}$
  3. $x^{2}E _{0}$
  4. -$x^{2}E _{0}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a uniform field E = E0, V = -integral(E dx) = -E0 * x + C. Since V(0) = 0, C = 0. Thus, V(x) = -E0 * x.

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

The potential $V$ is varying with x and y as $\displaystyle V = \dfrac{1}{2}(y^2-4x)$ volt. The field at $x = 1 m , y = 1 m$, is :

  1. $\displaystyle 2\hat{i}+\hat{j} \ Vm^{-1}$
  2. $\displaystyle -2\hat{i}+\hat{j} \ Vm^{-1}$
  3. $\displaystyle 2\hat{i}-\hat{j} \ Vm^{-1}$
  4. $\displaystyle -2\hat{i}+2\hat{j} \ Vm^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\displaystyle E _x = -\dfrac{dV}{dx} = -\dfrac{1}{2} [-4] = 2$

 $\displaystyle E _y = -\dfrac{dV}{dy} = -\dfrac{1}{2}[2y] = - y = -1$

 $\displaystyle \therefore \vec {E} = E _x \hat{i}+E _y \hat{j}=2\hat{i}-1\hat{j}$

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

The electric potential $V$ at any point $(x,y,z)$ in space is given by $V=4x^2$ volt. The electric field at $(1,0,2)$m in $Vm^{-1}$ is

  1. $8$, along the positive x-axis
  2. $8$, along the negative x-axis
  3. $16$, along the x-axis
  4. $16$, along the z-axis
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

E = -dV/dx. V = 4x^2, so E = -d(4x^2)/dx = -8x. At x = 1, E = -8 V/m. The negative sign indicates the direction is along the negative x-axis.

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

An electric field is expressed as $\displaystyle \vec{E} = 2\hat{i} + 3 \hat{j}$. Find the potential difference $(V _A - V _B)$ between two points $A$ and $B$ whose position vectors are given by $\displaystyle \vec r _A = \hat{i} + 2\hat{j}$ and $\displaystyle \vec r _B = 2\hat{i} + \hat{j}+3\hat{k}$ :

  1. $-1 V$
  2. $1 V$
  3. $2 V$
  4. $3 V$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$dV=-\vec E.\vec dx-\vec E.\vec dy$
$\Delta V=-\int Edx-\int Edy$
$\displaystyle V _B-V _A = -(\int _{1}^{2}2dx+\int _{2}^{1}3dy)$

$\displaystyle =-[2(2-1)+3(1-2)]$
$\displaystyle =-[2-3] = 1 V$
 Hence, $V _A-V _B = -1 V$

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

An infinite nonconducting sheet of charge has a surface charge density of $10^{-7}\ C/m^2$. The separation between two equipotential surfaces near the sheet whose potential differ by $5\ V$ is

  1. $0.88\ cm$
  2. $0.88\ mm$
  3. $0.88\ m$
  4. $5\times 10^{-7}\ m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The electric field of an infinite sheet is E = sigma / (2 * epsilon_0). Given sigma = 10^-7 C/m^2 and epsilon_0 = 8.85 * 10^-12 F/m, E = 10^-7 / (2 * 8.85 * 10^-12) = 10^5 / 17.7 = 5649 V/m. Using V = E * d, d = V / E = 5 / 5649 = 0.000885 m = 0.88 mm.