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 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.

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 is given as a function of distance by $V=(5x^2+10x-4)volt$, where x is in metre. Value of electric field at $x=1m$ is :

  1. $-23 V/m$
  2. $11 V/m$
  3. $6 V/m$
  4. $-20 V/m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given, $V=5x^2+10x-4$

or $\dfrac{dV}{dx}=10x+10$
The field, $E=-\dfrac{dV}{dx}=-(10x+10)$

At $x=1,  E=-(10+10)=-20 V/m$
So option D is correct. 

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 some charges situated on the x-axis is given by $V(x)=20/(x^2-4)volt$
The electric field E at $x=4\mu m$ is given by :

  1. $(10/9)volt /\mu m$ and in the $+ve$ x direction
  2. $(5/3)volt /\mu m$ and in the $-ve$ x direction
  3. $(5/3)volt /\mu m$ 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
A Correct answer
Explanation

Given, $\displaystyle V(x)=\dfrac{20}{x^2-4}$

Electric field , $\displaystyle E=-\dfrac{dV}{dx}=-\dfrac{-20}{(x^2-4)^2}(2x)=\dfrac{40x}{(x^2-4)^2}$

At $\displaystyle x= 4 \mu m,   E=\dfrac{40(4)}{(4^2-4)^2}=\dfrac{160}{144}=(10/9)  volt/\mu m$

Positive sign indicates that $\vec{E}$ is in the +ve 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

A and B are two points in an electric field. If the work done in carrying $4.0 C$ of electric charge from A to B is $16.0 J$, the potential difference between A and B is :

  1. $zero$
  2. $2.0 V$
  3. $4.0 V$
  4. $16.0 V$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The work done, $W _{A\rightarrow B}=q\int _{V _A}^{V _B}dV=q(V _B-V _A)$
or $16=4(V _B-V _A) \Rightarrow V _B-V _A=4  V$

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

Determine the electric field strength vector if the potential of the field depends on x, y coordinates as $V = a (x^2 - y^2)$, where a is a constant.

  1. $\vec{E} = - 2a(x\widehat{i} - y\widehat{j})$
  2. $\vec{E} = - a(x\widehat{i} - y\widehat{j})$
  3. $\vec{E} = - \dfrac{a(x\widehat{i} - y\widehat{j})}{2}$
  4. $\vec{E} = - \dfrac{a(x\widehat{i} - y\widehat{j})}{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ \vec E = - \triangledown V $   ( negative of gradient of V)

$\vec E = - (\dfrac{dV}{dx} \hat i + \dfrac{dV}{dy} \hat j )=-a(2x \hat i +2y \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

Determine the electric field strength vector if the potential of the field depends on x, y coordinates as $V = axy$ , where $a$ is a constant.

  1. $\vec{E} = -a(y\widehat{i} + y\widehat{j})$
  2. $\vec{E} = -a(x\widehat{i} + y\widehat{j})$
  3. $\vec{E} = -a(x\widehat{i} + x\widehat{j})$
  4. $\vec{E} = -a(y\widehat{i} + x\widehat{j})$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$ \vec E = - \triangledown V $   ( negative of gradient of V)

$\vec E = - (\dfrac{dV}{dx} \hat i + \dfrac{dV}{dy} \hat j )=-a(y \hat i + x \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 existing in space is $V(x, y, z) = A (xy+ yz + zx)$. Find the expression for the electric field :

  1. $-A{(x + Z) \widehat{i} + (y + Z) \widehat{j} + (x + y) \widehat{k}}$
  2. $-A{(y + Z) \widehat{i} + (x + Z) \widehat{j} + (x + y) \widehat{k}}$
  3. $-Ax{ \widehat{i} +y \widehat{j} + Z\widehat{k}}$
  4. $-A{(x+y) \widehat{i} + (x + y) \widehat{j} + (x + y-2Z) \widehat{k}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$ \vec E = - \triangledown V = -A[(y+z) \hat i + ( z+x) \hat j +( y+x) \hat k ] V/m $

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

At a certain distance from a point charge, the field intensity is 500 V/m and the potential is 3000 V. The distance and the magnitude of the charge respectively are :

  1. 6 m and 6 $\mu $C
  2. 4 m and 2 $\mu$C
  3. 6 m and 4 $\mu$C
  4. 6 m and 2 $\mu$C
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The electric field at distance d due to point charge q is $E=kq/d^2 $ and potential $V=kq/d$ 

so, $E=V/d $ or $ d=\dfrac{V}{E}=\dfrac{3000}{500}=6 m$

since, $V=kq/d $

or $3000=9\times 10^9\times \dfrac{q}{6} $

or $q=2\times 10^{-6} C=2 \mu C$