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

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

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 electric potential is $V (x, y, z) = Axy - Bx^2$ $+Cy$, where $A, B\ and\ C$ are positive constants. Calculate the $x, y\ and\ z$ components of the electric field.

  1. $E _x = - Ax, E _y = -Ay, E _z = 0$
  2. $E _x = - Ax + 2Bx, E _y = -Ay -C, E _z = 0$
  3. $E _x = - Ay + 2Bx, E _y = -Ax -C, E _z = 0$
  4. $E _x = - Ay, E _y = -Ax, E _z = 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here, $V(x,y,z)=Axy-Bx^2+Cy$


So, $E _x=-\dfrac{V}{dx}=-[Ay-2Bx]=2Bx-Ay$;


$E _y=-\dfrac{dV}{dy}=-[Ax+C]=-Ax-C$ and 

$E _z=-\dfrac{dV}{dz}=0$

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

Potential difference between centre and surface of the sphere of radius R and uniform volume charge density $\rho$ within it will be :

  1. $\displaystyle \dfrac{\rho R^2}{6 \varepsilon _0}$
  2. $\displaystyle \dfrac{\rho R^2}{4 \varepsilon _0}$
  3. $\displaystyle \dfrac{\rho R^2}{3 \varepsilon _0}$
  4. $\displaystyle \dfrac{\rho R^2}{2 \varepsilon _0}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Gauss's law the electric field inside the sphere , $E.4\pi r^2=\dfrac{q}{\epsilon _0}=\rho\dfrac{(4/3)\pi r^3}{\epsilon _0}$

or $E=\dfrac{\rho r}{3\epsilon _0}$

Potential difference between surface and center is $V=-\int _R^0 E.dr=-\int _R^0 \dfrac{\rho r}{3\epsilon _0} dr=\dfrac{\rho R^2}{6\epsilon _0}$

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 exists in x-y plane. The potential of points A (-2m, 2m), B(+2m, 2m) and C(2m, 4m) are 4 V, 16V and 12 V respectively. The electric field is :

  1. $(4\widehat{i} + 5 \widehat{j}) V/m$
  2. $(3\widehat{i} + 4 \widehat{j}) V/m$
  3. $-(3\widehat{i} + 4 \widehat{j}) V/m$
  4. $(3\widehat{i} - 4 \widehat{j}) V/m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let equation of potential be $ax+by+c$ where $(x,y)$ are the co-ordinates of the point in $x, y$ plane.
so, $a(2)+b(2)+c=4$
$a(-2)+b(2)+c=16$
$a(2)+b(4)+c=12$
Solving above equations, we get $a=-3;b=4;c=2$
so equation of potential is $V=-3x+4y+2$ 
Now the electric field is $\vec E=(-dV/dx)\vec i+(-dV/dy)\vec j=3\vec i-4\vec j$
Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

In a certain region of space, the electric potential is $V (x, y, z) = Axy - Bx^2$ $+Cy$, where $A, B\ and\ C$ are positive constants. At which points is the electric field equal to zero?

  1. $x = +C/A, y = +BC/A$ $^2$, any value of $z$
  2. $x = +C/A, y = +2BC/A$ $^2$, any value of $z$
  3. $x = -C/A, y = -2BC/A$ $^2$, $z=0$
  4. $x = -C/A, y = -2BC/A$ $^2$, any value of $z$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$ \vec E = - \triangledown V = - [ ( Ay - 2Bx) \hat i + ( Ax + C) \hat j ] $


$ \vec E = 0$  at, 

$ E _y = 0 \Rightarrow Ax + C= 0 \Rightarrow x = -C/A $

$ E _x= 0 \Rightarrow Ay - 2Bx= Ay + 2BC/A =0 \Rightarrow y = -2BC/A^2 $

$ E _z = 0 $  everywhere . 

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 $V(x, y, z) = A (xy+ yz + zx)$. If A is $10$ SI units, find the magnitude of the electric field at $(1 m, 1 m, 1 m)$ :

  1. $20 \sqrt 2$ N/C
  2. $20 \sqrt 3$ N/C
  3. $10 \sqrt 3$ N/C
  4. $20 $ N/C
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 ] = -10[2 \hat i + 2 \hat j + 2 \hat k ] = 20\sqrt{3} N/C$

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 at a point (x, y) in the x-y plane is given by V = - Kxy. The field intensity at a distance r in this plane, from the origin is proportional to :

  1. $r^2$
  2. $r$
  3. $1/r$
  4. $1/r^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let the $x-y$ coordinates of the point at distance $r$ from the origin be given as $x=rcos\theta$, $y=rsin\theta$
Potential is given as $V=-Kxy$
Now, Electric field intensity is $\vec E=(-dV/dx)\vec i+(-dV/dy)\vec j=K(y\vec i+x\vec j)=K(rsin\theta\vec i+rcos\theta\vec j)=Kr(sin\theta\vec i+cos\theta\vec j)\propto r$
So electric field potential is proportional to $r$.
Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

Find out the relationship between the electric field and electric potential include which of the following statement?

I. If the electric field at a certain point is zero, then the electric potential at the same point is also zero.

II. The electric potential is inversely proportional to the strength of the electric field.

III. If the electric potential at a certain point is zero, then the electric field at the same point is also zero.

  1. I only

  2. II only

  3. I and II only

  4. I and III only

  5. None of the above
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

If electric field at some point is zero, it is not necessary for the electric potential to be the same. Consider the mid point of line joining to equal charges of same sign. Field there is zero, but potential is finite and positive.


Electric field strength is given by $E=\dfrac{kQq}{r^2}$
Electric potential is given by $V=\dfrac{dQq}{r}$
Clearly they are not inversely proportional to each other.

If electric potential at some point is zero, it is not necessary for the electric field to be the same. Consider the mid point of line joining to equal charges of opposite signs. Potential there is zero, but electric field exists.

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 (in $V/m$) is given by $E=10x^3$. Determine the potential difference, in volts, between $x=0m$ and $x=3m$.

  1. $202.5$
  2. $100$
  3. $20$
  4. $250$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that, $E=-\dfrac{dV}{dx}$


$V=-\int _0^3 Edx=-\int _0^3 10x^3 dx=-10\times \dfrac{3^4}{4}=202.5 $