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 static electricity exchange of electrons methods used for charging electric charge

A hollow metal sphere, Sphere A, sits on an insulating stand. Sphere A has a diameter of 4 inches, and a net charge of magnitude $Q _0$. A second hollow metal sphere, Sphere B, also sits on an insulating stand, but has a diameter of 8 inches and zero net charge. The two spheres are brought close so that they touch, then they are separated.
In terms of $Q _0$, what is the final charge on Sphere A?

  1. $\cfrac{Q _0}{5}$
  2. $\cfrac{Q _0}{4}$
  3. $\cfrac{Q _0}{2}$
  4. $Q _0$
  5. $4Q _0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$R _B=2R _A$


Let final charge on spheres be $Q _A$ and $Q _B$.

On touching the spheres charge density becomes same.

And, charge density, $\sigma=\dfrac{Q}{4\pi R^2}$

$\implies Q\propto R^2$ for same $\sigma$

Hence, $\dfrac{Q _B}{Q _A}=\dfrac{R _B^2}{R _A^2}=2^2$

$\implies Q _B=4Q _A$

And, total charge$=Q _B+Q _A=Q _o$

$\implies 4Q _A+Q _A=Q _o$

$\implies Q _A=\dfrac{Q _o}{5}$

Answer-(A)

Multiple choice physics some natural phenomena types of charges and their interaction transfer of charges charging and discharging

Rub an empty ball pen refill on a polythene sheet and hold it on top of small pieces of paper. What will be your observation?

  1. Pieces of paper are attracted.

  2. Pieces of paper are repelled.

  3. Pieces of paper are neither attracted nor repelled.

  4. Pieces of paper are either attracted or repelled.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When refill is rubbed on polythene it acquires small electric charges. Due to presence of these charges, pieces of paper gets attracted.

Multiple choice physics some natural phenomena types of charges and their interaction transfer of charges charging and discharging transfer of charge

Inflate two balloons. Hang them in such a way that they do not touch each other. Rub both balloons with woollen cloth and release them. Then:

  1. both balloons attract each other.

  2. both balloons repel each other.

  3. both balloons remain at same position.

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

The balloons will be seen to repel each other as both are charged with same polarity.

Multiple choice physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

A small charged ball of mass m and charge q is suspended from the highers point of a ring of radius R by means of an insulated code of negligible mass.The ring is made of a rigid wire of negligible cross-section and lies in a vertical plane.On the ring, there is uniformly distributed charge Q of the same as that of q .determine the length of the cord so as the equilibrium position of the ball lies on the symmetry axis ,perpendicular to the plane of the ring. 

  1. $\left( \cfrac { 2kQqR }{ mg } \right) ^{ 1/3 }$
  2. $\left( \cfrac { kQqR }{ mg } \right) ^{ 1/3 }$
  3. $\left( \cfrac { kQqR }{ 2mg } \right) ^{ 1/3 }$
  4. $\left( \cfrac { kQqR }{ mg } \right) ^{ 3 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice motion of charged particle in magnetic field and electric field moving charges and magnetism magnetic effects of current and magnetism physics

A uniform electric field 'E' is directed towards positive X-axis. If at X=0, the electric potential is zero, then the potential at $X=+X _0,$ would be 

  1. $\dfrac{E}{X _0}$
  2. $\dfrac{-E}{X _0}$
  3. $-EX _0$
  4. $EX _0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that electric potential $V=\dfrac { dE }{ dx } $

Since $X=x+{ x } _{ 0 }$  so,
$V=\dfrac { E }{ { x } _{ 0 } } $

Multiple choice motion of charged particle in magnetic field and electric field moving charges and magnetism magnetic effects of current and magnetism physics

Three equal charges, each having a magnitude of $ 4 \mu C$ , are placed at the three corners of a right-angled triangle of sides $6 cm, 8 cm$ and $10 cm.$ The force on the charge at the right-angle corner will be

  1. $ 11.5 N $
  2. $23 N $
  3. $46 N $
  4. $230 N $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Two forces are,

${F _1} = \dfrac{{k{Q _1}{Q _2}}}{{{R _1}^2}}$

$ = \dfrac{{9 \times {{10}^9} \times {{\left( {4 \times {{10}^{ - 6}}} \right)}^2}}}{{{{\left( {6 \times {{10}^{ - 2}}} \right)}^2}}}$

$ = 10{\rm{N}}$

${F _2} = \dfrac{{k{Q _1}{Q _2}}}{{{R _2}^2}}$

$= \dfrac{{9 \times {{10}^9} \times {{\left( {4 \times {{10}^{ - 6}}} \right)}^2}}}{{{{\left( {8 \times {{10}^{ - 2}}} \right)}^2}}}$

$= \dfrac{{360}}{{64}} = 5.62{\rm{N}}$

Resultant force at the right angle vertex is,

$F = \sqrt {\left( {{F _1}^2 + {F _2}^2} \right)} $

$= \sqrt {{{10}^2} + {{5.62}^2}} $

$ = \sqrt {131.56} $

$= 11.5{\rm{N}}$

Multiple choice motion of charged particle in magnetic field and electric field moving charges and magnetism magnetic effects of current and magnetism physics

If uniform electric field $\vec{E} = E _0 \hat{i} + 2E _0 \hat{j}$ where $E _0$ is a constant, exists in a region of space and at (0, 0) the electric potential V is zero, then the potential at $(x _0, 0)$ will be 

  1. zero
  2. $-E _0 x _0$
  3. $-2 \, E _0 x _0$
  4. $-\sqrt{5} E _0 x _0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The electric field is given by E = E0 i + 2E0 j. The potential difference between two points is V(x,y) - V(0,0) = - integral (E dot dl). Integrating along the x-axis from 0 to x0 gives V(x0,0) - 0 = - E0 * x0, resulting in -E0 x0.

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 $ne=6.40 \times 10^{-19} C$
$n=\dfrac{6.40 \times 10^{-19} }{1.6\times 10^{-19}}=4$