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 parallel plate capacitor electrostatic potential and capacitance electrostatics physics

Find the potential at a point due to a positive charge of $100\mu C$ at a distance of $10\ m$ in a medium of dielectric constant $9$.

  1. $10^{7}V$.
  2. $10^{4}V$.
  3. $10^{5}V$.
  4. $10^{6}V$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Potential V = (1 / (4 * pi * epsilon_0 * K)) * (Q / r). Given Q = 100 * 10^-6 C, r = 10 m, K = 9. V = (9 * 10^9 / 9) * (100 * 10^-6 / 10) = 10^9 * 10^-5 = 10^4 V.

Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

Among two discs $A$ and $B$, first have radius $10\ cm$ and charge ${10}^{-6}\ \mu C$ and second have radius $30\ cm$ and charge ${10}^{-5}C$. When they are touched, charge on both ${q} _{A}$ and ${q} _{B}$ respectively will be :

  1. ${q} _{A}=2.75\mu C,{q} _{B}=3.15\mu C$
  2. ${q} _{A}=1.09\mu C,{q} _{B}=1.53\mu C$
  3. ${q} _{A}={q} _{B}=5.5\mu C$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

Two metal pieces having a potential difference of 800 V are 0.02 m apart horizontally. A particle of mass $1.96\times 10^{-15}kg$ is suspended in equilibrium between the plates. If e is the elementary charge, then charge on the particle is

  1. 8

  2. 6

  3. 0.1

  4. 3
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Force due to gravity on the particle is $F _g=mg$
Force due to field E between the metal plates is $\displaystyle F _e=qE=(ne)\frac{V}{d}$.
In equilibrium, $\displaystyle F _g=F _e \Rightarrow mg=(ne)\frac{V}{d}$


$\displaystyle \therefore n=\dfrac{mgd}{eV}$

$=\dfrac{1.96\times 10^{-15}\times 9.8\times 0.02}{1.6\times 10^{-19}\times 800}=3$

Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

Which of the following is true about field between parallel charged plates?

  1. It is strongest between the plates

  2. It is strongest near the positive plate

  3. It is strongest near the negative plate

  4. The field is constant between the plates

  5. The field is variable, therefore the strong point also varies
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The field is constant between the plates , because we dealing with a system of two charged plates.we know the electric field between two charged plates (suppose both have positive charge) is given by 

                   $E=\frac{1}{2\varepsilon _{0}}\left(\sigma _{1}-\sigma _{2}\right)$         where $\sigma=$ surface charge density
we can see that electric field doesn't depend upon distance from plates , therefore it is constant.

Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

A thunder cloud and the earth's surface may be regarded as a pair of charged parallel plates separated by a distance $h$ and the capacitance of the system is $C$. When a flash of mean current '$i$' occurs for a time duration '$t$', the electric field strength between the cloud and earth is:

  1. $\dfrac { it }{ C } $
  2. $Cit$
  3. $\dfrac { it }{ Ch } $
  4. $\dfrac { Cit }{ h } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total charge accumulation in time duration $t$ is:

$Q = it$

From definition of capacitance, voltage induced due to this charge is:
$V = \cfrac{Q}{C}$
$V = \cfrac{it}{C}$

Electric field is defined as negative of gradient of potential. Hence,
$E = -\cfrac{dV}{dr}$
$\left| E \right| = \cfrac{V}{h}$
$\left| E \right| = \cfrac{it}{Ch}$

Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

Two point charges $17.7 \mu c$ and $-17,7 \mu c$ separated by a very small distance, are kept inside a large hollow metallic sphere. Electric flux emnating through the sphere is :

  1. $ 2 \times 10^6$Vm
  2. $- 2 \times 10^6$Vm

  3. Zero
  4. $ 4 \times 10^6$Vm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since net charge is zero

$\therefore$ Net flux is zero.
Hence,
option $C$ is correct answer.

Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

In 1909, Robert Millikan was the first to find the charge of an electron in his now-famous oil-drop experiment. In that experiment, tiny oil drops were sprayed into a uniform electric field between a horizontal pair of oppositely charged plates.The drops were observed with a magnifying eyepiece, and the electric field was adjusted so that the upward force on some negatively charged oil drops was just sufficient to balance the downward force of gravity. That is, when suspended, upward force qE just equaled mg. Millikan accurately measured the charges on many oil drops and found the values to be whole number multiples of $1.6  \times 10^{-19} C$ the charge of the electron. For this, he won the Nobel prize. Extra electrons on this particular oil drop (given the presently known charge of the electron) are :

  1. $4$
  2. $3$
  3. $5$
  4. $8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that electrostatic force is just balancing gravitational force.
$qE = mg$   

$q \times 1.68 \times 10^5 = 1.08 \times 10^{-14} g$
Charge of drop, $q = 6.40 \times 10^{-19} C$
Charge of an electron $=1.6\times 10^{-19}$
Let the no. of electrons on the drop is $n$.
Then $ne=6.40 \times 10^{-19} C$
$n=\dfrac{6.40 \times 10^{-19} }{1.6\times 10^{-19}}=4$

Multiple choice physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

Two infinite linear charges are placed parallel to each other at a distance 0.1 m from each other. if the linear charge density on each is 5 $ 5 \mu \ C /m $ , then the force acting on a unit length of each linear charge will be

  1. 2.5 N/m

  2. 3.25 N/m

  3. 4.5 N/m

  4. 7.5 N/m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The force per unit length between two infinite parallel wires is F/L = (mu_0 * lambda_1 * lambda_2) / (2 * pi * d). Using mu_0 = 4 * pi * 10^-7, lambda = 5 * 10^-6 C/m, and d = 0.1 m, the calculation yields 4.5 N/m.

Multiple choice physics magnetic fields and electromagnetism magnetic flux density magnetic flux electromagnetic induction

A charge q is placed at the centre of a cylinder of radius R and length 2R. Then electric flux through the curved surface of the cylinder is 

  1. $\cfrac { q }{ 2 { \epsilon } _{ 0 } } $
  2. $\cfrac { q }{ 4 { \epsilon } _{ 0 } } $
  3. $\cfrac { q }{ \sqrt { 2 } { \epsilon } _{ 0 } } $
  4. $\cfrac { q }{ 2\sqrt { 2 } { \epsilon } _{ 0 } } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics magnetic fields and electromagnetism magnetic flux density magnetic flux electromagnetic induction

A sphere of radius $R$ and charge $Q$ is placed inside an imaginary sphere of radius $2R$. Whose center coincides with the given sphere. The flux related to the imaginary sphere is:

  1. $\dfrac {Q}{\in _{0}}$
  2. $\dfrac {Q}{2\in _{0}}$
  3. $\dfrac {4Q}{\in _{0}}$
  4. $\dfrac {2Q}{\in _{0}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

According to Gauss's Law, the total electric flux through any closed surface is equal to the enclosed charge divided by epsilon_0. Since the inner sphere with charge Q is entirely enclosed by the imaginary sphere, the flux is Q/epsilon_0.

Multiple choice physics magnetic fields and electromagnetism magnetic flux density magnetic flux electromagnetic induction

The electric field in a certain region is $\left( 10\hat { i } +5\hat { j }  \right) \times { 10 }^{ 4 }N/C$. What is the flux due to this field over an area of $\left( 3\hat { i } +3\hat { j }  \right) \times { 10 }^{ -2 }{ m }^{ 2 }$ in ${ Nm }^{ 2 }/C?$

  1. $4.5\times { 10 }^{ 3 }$
  2. $3.5\times { 10 }^{ 3 }$
  3. $2.5\times { 10 }^{ 3 }$
  4. $1.5\times { 10 }^{ 3 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Electric flux is the dot product of the electric field vector and the area vector. Phi = (10i + 5j) * 10^4 * (3i + 3j) * 10^-2 = (30 + 15) * 10^2 = 45 * 10^2 = 4.5 * 10^3 Nm^2/C.

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

Two parallel plates have equal and opposite charge. When the space between them is evacuated. the electric field between the plates $2 \times {10^5}\,V/m.$ When the space is filed with dielectric the electric field becomes ${10^5}\,V/m$ The dielectric constant of he dielectric material is 

  1. $2$
  2. $4$
  3. $5$
  4. $9$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Dielectric constant$:-$

$K = \dfrac{{{E _0}}}{E}$
$ = \dfrac{{2 \times {{10}^5}}}{{1 \times {{10}^5}}} = 2$
Hence,
option $(A)$is correct answer