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 electric fields introduction to electrostatic force electric force charging and discharging

Two copper spheres, $A$ and $B$, are identical in all respect but A carries a charge of $-3 \mu C$ whereas $B$ Is charged to $+1 \mu C$. The two spheres are brought together until they touch and then separated by some distance. Which of the following statements is true concerning the electrostatic force $F$ between the spheres?

  1. $F = 0$ as one of the spheres is uncharged
  2. $F = 0$ as both the spheres are uncharged
  3. $F$ is attractive
  4. $F$ is repulsive.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Charge on $A^{+}$ sphere $=-3$ $uC$

Charge on $B$ sphere $=+1$ $uC$
When they are connected the charge get redistributed, i.e.
$\Rightarrow$ Charge on both sphere $=\dfrac{-3+1}{2}$ $uC=-1$ $uC$
So charge on either sphere $=-1$ $uC$
Therefore, After connecting charge on sphere $A=-1$ $uC$
After connecting charge on sphere $B=-1$ $uC$
Therefore, there will be a repulsive force between the sphere because of like charges.

Multiple choice physics electric fields introduction to electrostatic force electric force charging and discharging

The potential at a point $(x, 0, 0)$ is given by $V = \left(\dfrac{1000}{x} + \dfrac{1500}{x^2} + \dfrac{500}{x^3}\right)$. The intensity of the electric field at $x = 1$ will be

  1. $550 V/m$
  2. $55 V/m$
  3. $55000 V/m$
  4. $5500 V/m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$V=(\cfrac{1000}{x}+\cfrac{1500}{x^{2}}+\cfrac{500}{x^{3}})$
We know that, $E=\cfrac{-dv}{dr}$
So,
$E=\cfrac{-d(\cfrac{1000}{x}+\cfrac{1500}{x^{2}}+\cfrac{500}{x^{3}})}{dx}$   at $x=1$
$=-[\cfrac{(-1000)}{x^{2}}-\cfrac{2\times 1500}{x^{3}}-\cfrac{3\times 500}{x^{4}}] _{x=1}$
$=1000+3000+1500=5500\,V/m$
Multiple choice physics electric current, potential difference and resistance electric potential and potential difference potential difference current in electric circuits

A big hallow metal sphere $A$ is charged to $100$ volts and another smaller hollow sphere $B$ is charged to $50$ volts. If B is put inside $A$ and joined with a metallic wire, then the direction of charge flow:-

  1. is from $A$ to $B$
  2. is from $B$ to $A$

  3. to charge flows

  4. depends on the radii of spheres
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Charge flows from a higher potential to a lower potential. Since sphere A is at 100V and sphere B is at 50V, charge will flow from A to B until they reach the same potential.

Multiple choice physics electric current, potential difference and resistance electric potential and potential difference potential difference current in electric circuits

Two conducting parallel plates areseparated by a distance of 0.001$\mathrm { m } . \mathrm { A } 9 \mathrm { V }$battery is connected across the plates.Find out the electric field between the plates? 

  1. 9000$\mathrm { V } / \mathrm { m }$
  2. 900$\mathrm { Vim }$
  3. 9$\mathrm { V } / \mathrm { m }$
  4. .9$\mathrm { V } / \mathrm { m }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The electric field E between parallel plates is given by E = V / d. Given V = 9 V and d = 0.001 m, E = 9 / 0.001 = 9000 V/m.

Multiple choice various types of barometer fluids physics

Eight identical spherical mercury drops charged to a potential of $20V$ each are coalesced into a single spherical drop.

  1. The internal Energy of the system remains the same.
  2. The new potential of the drop is $80V$

  3. Internal energy of the system decreases
  4. The potential remains the same i.e., $20V$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Potential of one small drop of mercury, 
$V=\displaystyle\frac{kq}{r}=20V$
Volume of big drop=volume of $8$ small drops
$\displaystyle\frac{4}{3}\pi R^3=8\times \frac{4}{3}\pi r^3\Rightarrow =2r$
$Q'=8q$
Potential of big drop,
$\displaystyle V'=\frac{kQ'}{R}=\frac{K\times 8q}{2r}=\frac{4kq}{r}=4\times 20=80V$
Hence, option $B$ is the correct answer.

Multiple choice physics static electricity explaining static electricity charging by induction electric charges and fields

Two bodies are changed by rubbing one against the other. During the process, one becomes positively charged while the other becomes negatively charged. Then,

  1. mass of each body remains unchanged.

  2. mass of each body changes marginally.

  3. mass of each body changes slightly and hence the total mass.

  4. mass of each body changes slightly but the total mass remains the same.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The transfer of electrons from one body to the result in redistribution of charges.
Hence, no. of electrons given by one body $=$ Number of electrons obtained by the other.
$\therefore$ Mass of negatively charged body slightly increases.
Whereas the total mass of the system remain the same.
Multiple choice physics static electricity explaining static electricity charging by induction electric charges and fields

A hollow metallic sphere is charged. Inside the sphere

  1. The potential is zero but the electric field is finite

  2. The electric field is zero but the potential is finite

  3. Both the electric field and the potential are finite

  4. Both the electric field and the potential are zero
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a charged hollow metallic sphere, the electric field inside is zero because the charges reside on the outer surface. However, the potential inside is constant and equal to the potential at the surface, which is finite.

Multiple choice physics static electricity explaining static electricity charging by induction electric charges and fields

If a body is charged by rubbing it, its weight _________.

  1. Remains precisely constant

  2. Increases slightly

  3. Decreases slightly

  4. May increase or decrease slightly
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If a body is charged by rubbing it, then it may lose or gain electrons. Since electrons have a mass of $(9.1\times 10^{-31} :Kg)$. So, a slight weight may increase or decrease slightly.

Multiple choice physics static electricity explaining static electricity charging by induction electric charges and fields

A glass rod when rubbed with silk cloth, acquires a charge of $1.6\times 10^{-11}C$, then the charge on silk cloth will be:

  1. $-3.2\times 10^{-11}C$
  2. $-2.4\times 10^{-13}C$
  3. $-1.6\times 10^{-13}C$
  4. $-1.6\times 10^{-11}C$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When glass rod is rubbed with silk, electrons move from rod to silk.


Since silk gets electrons it becomes negatively charged and the number of electrons gained by silk is same as that lost by rod.

Hence magnitude of charge on silk is same as that on rod.

Hence charge on silk$=-1.6\times 10^{-11}C$ 

Answer-(D)

Multiple choice physics energy : forms and sources renewable and non-renewable resources renewable and non-renewable sources of energy production of electricity from solar energy

A mercury drop of water has potential 'V' on its surface. $1000$ such drops combine to form a new drop. Find the potential on the surface of the new drop.

  1. V
  2. $10$V
  3. $100$V
  4. $1000$V
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Potential of single smaller Hg drop of water =$\dfrac{Kq}{r}$=V-----(1)

and we know potential of n smaller Hg drop of water $=\dfrac{Knq}{R}$------(2)
Volume of 1000 small drops = volume of larger drop
$1000\dfrac{4}{3}\pi {r}^{3}=\dfrac{4}{3}\pi {R}^{3}$
Thus radius of bigger droplet $R=10r$
 Now putting the value of R in equation 2 we get
Potential of larger drop$=\dfrac{1000Kq}{10r}$=$100V$  

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 ratio of electric force $ ( F _e ) $ to gravitational force acting between two electrons will be:

  1. $ 1 \times 10^{36} $
  2. $ 2 \times 10^{39} $
  3. $ 2.5\times 10^{39} $
  4. $ 3 \times 10^{39} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The ratio of electrostatic force to gravitational force between two electrons is approximately 4.17 * 10^42. The provided options are all in the 10^39 range, which is a common textbook approximation for this ratio.

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

Electric potential at ( x, y, z ) is given as $V$= $- x ^ { 2 } y \sqrt { z }$ Find the electrical field at (2 ,1, 1)

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

The electric field is the negative gradient of potential: E = -∇V. Computing partial derivatives: ∂V/∂x = -2xy√z, ∂V/∂y = -x²√z, ∂V/∂z = -x²y/(2√z). At point (2,1,1): Ex = -2(2)(1)(1) = -4, Ey = -(4)(1) = -4, Ez = -(4)(1)/(2×1) = -2. Therefore E = -(-4i - 4j - 2k) = 4i + 4j + 2k. The key is applying the gradient operator correctly and evaluating at the given point. Note that option D has a typo (should be î, not 1̂) but is clearly the intended 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 and the electric potential at a point inside a shell are E and V respectively. Which of the following is correct?

  1. If $E=0$, V must be zero.
  2. If $V=0$, E must be zero.
  3. If $E\neq 0$, V cannot be zero.

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In a shell $E=0 $ but $V \neq 0$.
Along the equatorial line of a dipole, $V=0 $ but $E  \neq 0$.