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

A charge of $6.76$ $\mu$C in an electric field is acted upon by a force of $2.5 N$. The potential gradient at this point is :

  1. $3.71 \times 10^{15} Vm^{-1}$
  2. $-3.71 \times 10^{12} Vm^{-1}$
  3. $3.71 \times 10^{10} Vm^{-1}$
  4. $-3.71 \times 10^{5} Vm^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$F=qE \Rightarrow E=\dfrac{F}{q}=\dfrac{2.5}{6.76\times 10^{-6}}=3.7\times 10^5$

The potential gradient is the electric field i.e, $E=-\nabla V=-3.7 \times 10^5 $

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 $20$ V/cm
  2. may be equal to $20$ V/cm
  3. may be greater than $20$ V/m
  4. may be less than $20$ V/cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As the electric potential decreases uniformly so the electric field is uniform along the x axis.

$E=-\dfrac{dV}{dx}=\dfrac{120-80}{1-(-1)}=20 $V/m. It must be equal to 20 V/m at the origin.

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 in a region is directed outward and is proportional to the distance r from the origin. Taking the electric potential at the origin to be zero,

  1. it is uniform in the region

  2. it is proportional to r
  3. it is proportional to r$^2$

  4. it increases as one goes away from the origion
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Electric field is directly propotional to $r$, therefore
$E=kr$
We know that
$\ V=-\int _{ 0 }^{ r }{ \overrightarrow { E. } \overrightarrow { dr }  } $
which gives 
$V=-\dfrac { k{ r }^{ 2 } }{ 2 } $ since V at $r=0$ is $0$
Hence V is proportional to ${ r }^{ 2 }$

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 uniform electric field of $20$ NC$^{-1}$ exists along the x-axis in space. The potential difference V$ _B-$V$ _A$ for the point A $=$ $(4 m, 2m)$ and B $=$ $(6m, 5m)$ is:

  1. $20$ $\sqrt{13}$ V
  2. $- 40 V$

  3. zero V

  4. none of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here, $\vec{E}=20 \hat i \Rightarrow E _x=20$
Potential difference ,$V _B-V _A=-\int _A^B Edr=-\int _4^6E _x dx=-20\int _4^6 dx=-20(6-4)=-40  V$

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 electric field at the origin is along the positive X-axis. A small circle is drawn with the centre at the origin cutting the axes at points A, B, C and D having coordinates $(a, 0), (0, a), (-a, 0), (0, -a)$ respectively. Out of the given points on the periphery of the circle, the potential is minimum at :

  1. A

  2. B

  3. C

  4. D
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The relation between electric field and potential is given by $\vec{E} = \displaystyle -\frac{\partial V}{\partial x}\hat{i} -\frac{\partial V}{\partial y}\hat{j} $

Given that, at origin, electric field is along positive x-axis.
Thus, $\displaystyle \frac{\partial V}{\partial x} < 0$ and $\displaystyle \frac{\partial V}{\partial y} = 0$
Thus, $V$ decreases in the positive x-direction and remains constant in y-direction.
Hence, minimum $V$ occurs at $(a,0)$ i.e., $A$

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

It is found that air breaks down electrically, when the electric field is $  3 \times 10^{6} \mathrm{V} / \mathrm{m} .  $ What is the potential to which a sphere of radius $1  \mathrm{m}  $ can be raised, before sparking takes place?

  1. $ V=10^{6} \mathrm{V} $
  2. $ V=2 \times 10^{6} \mathrm{V} $
  3. $ V=3 \times 10^{6} \mathrm{V} $
  4. $ V=4 \times 10^{6} \mathrm{V} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a charged conducting sphere, the electric field at the surface is E = kQ/R^2 and the potential is V = kQ/R. Therefore, V = E * R. Given E = 3 * 10^6 V/m and R = 1 m, V = 3 * 10^6 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

In moving from A to B along an electric field line, the wok done by the electric field on an electron is $6.4 \times 10^{-19}$ J. If $\phi _1$ and $\phi _2$ are equipotential surfaces, then the potential difference $V _b-V _A $ is

  1. -4V

  2. 4V

  3. zero

  4. 6.4 V
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Work done by the electric field is W = q * (V_A - V_B). For an electron, q = -1.6 * 10^-19 C. Given W = 6.4 * 10^-19 J, we have 6.4 * 10^-19 = -1.6 * 10^-19 * (V_A - V_B), which simplifies to V_A - V_B = -4 V, or 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

The electric potential in a certain region along the x-axis varies with x according to the relation $V(x) = 5 - 4x^2$. Then, the correct statement is :

  1. the potential difference between the points $x =1$m and $x=2$m is $12$ Volt
  2. the force experienced by a Coulomb of charge placed at $x =1$ m is $8$ Newton

  3. the electric field components along Y and Z direction are zero

  4. all of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$V(x)=5-4x^2$

$V(1)=5-4=1 V,  V(2)=5-4(2^2)=-11 V$

Potential difference between $x=1 m$ and $x=2 m$ is $V _{12}=V _1-V _2=1-(-11)=12 V$

here, $E _x=-\dfrac{dV}{dx}=8x,  E _y=-\dfrac{dV}{dy}=0$ and $E _z=-\dfrac{dV}{dz}=0$

The electric force on $1$ coulomb charge at $x=1$ is $F=qE _x=1(8)=8 N$

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
B 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 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}$