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 dielectric substances and polarization dielectrics and polarisation electrostatic potential and capacitance electrostatics physics

Two charges when kept at a distance of $1 m$ apart in vacuum have some force of repulsion. If the force of repulsion between these two charges be same, when placed in an oil of dielectric constant $4$, the distance of separation is 

  1. $0.25m$
  2. $0.4m$
  3. $0.5m$
  4. $0.6m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The electrostatic force between two charges in a medium of dielectric constant K is given by F = (1 / (4 * pi * epsilon_0 * K)) * (q1 * q2 / r^2). For the force to remain the same when placed in an oil of dielectric constant 4, the effective distance r' must be r / sqrt(K). Since r = 1 m and K = 4, r' = 1 / sqrt(4) = 0.5 m.

Multiple choice dielectric substances and polarization dielectrics and polarisation electrostatic potential and capacitance electrostatics physics

Surface charge density of a thin disc having radius $R$ varies with distance from centre as $\sigma =\sigma _{0}\dfrac {R}{r}(r\neq 0) $ then total charge of disc is

  1. $2\pi \sigma _{0}R^{3}$
  2. $\sqrt {2}\pi \sigma _{0}R^{2}$
  3. $2\sqrt {2}\pi \sigma _{0}R^{2}$
  4. $2\pi \sigma _{0}R^{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The total charge of a thin disc with radially varying surface charge density sigma = sigma_0 * (R / r) is found by integrating dq = sigma * 2 * pi * r * dr from r = 0 to r = R. Substituting sigma gives the integral of 2 * pi * sigma_0 * R * dr, which evaluates to 2 * pi * sigma_0 * R^2.

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A simple pending of length l has a bob of mass m, with a charge q on it . A  vertical sheet of charge, with surface charge density $\sigma $ passes string makes an angle $\theta $ with the vertical , then 

  1. $\quad tan\theta =\dfrac { \sigma q }{ 2{ \epsilon } _{ 0 }mg } $
  2. $\quad tan\theta =\dfrac { \sigma q }{ { \epsilon } _{ 0 }mg } $
  3. $\quad cot\theta =\dfrac { \sigma q }{ 2{ \epsilon } _{ 0 }mg } $
  4. $\quad cot\theta =\dfrac { \sigma q }{ { \epsilon } _{ 0 }mg } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The electric field due to a sheet of charge is E = sigma / (2*epsilon_0). The force on the bob is qE. The angle theta satisfies tan(theta) = F_electric / F_gravity = (q*sigma / (2*epsilon_0)) / (mg).

Multiple choice physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

A charge $A$ of $+3 \ mC$ is placed at $k=0$ and a charge $B$ of $-5 \ mC$ at $k=40 \ mm.$ Where a third charge q be placed on the axis such that it experiences no force is

  1. $1.6 \times 10^{-1} \ m$ from $B$ outside
  2. $2.52 \times 10^{-1} \ m$ from $B$ outside
  3. $4.42 \times 10^{-1} \ m$ from $B$ outside
  4. $8.24 \times 10^{-1} \ m$ from $B$ outside.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a charge to experience no force, the electric fields from the two charges must cancel out. Since the charges have opposite signs, the null point must be outside the region between them, closer to the smaller magnitude charge (3 mC). Solving k(q1)/x^2 = k(q2)/(x+d)^2 leads to the position 1.6 x 10^-1 m from the -5 mC charge.

Multiple choice physics energy and its forms idea of energy introduction to work and energy work and energy

Work done charge of mass 2 Kg due to external force against electrostatics force is - 10 J if charge is displaced from A to B. Velocity of charge at point A is 4m/s and at B is 2m/s then find difference is electropotential energy $ (U _s - U _A) $

    • 10J
    • 10J
    • 21J
    • 2J
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By the work-energy theorem, the total work done equals the change in kinetic energy. The work done by the external force is -10 J. The change in kinetic energy is 1/2 * m * (v_B^2 - v_A^2) = 1/2 * 2 * (4 - 16) = -12 J. The change in potential energy is the negative of the work done by the conservative (electrostatic) force. Given the total work W_ext + W_elec = Delta K, we find Delta U = +10 J.

Multiple choice physics magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

Charge is uniformly distributed in  a space. The net flux passing through the surface of an imaginary cube of side''a'' in the spaceis $\phi $ the space is 0. The net flux passing through the surface of an imaginary sphere of radius ''a''- in the space will be:

  1. $\phi $
  2. $\dfrac { 3 }{ 4\pi } \phi $
  3. $\dfrac {2\pi }{ 3 } \phi $
  4. $\dfrac {4\pi }{ 3 } \phi $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

external flux of a surface is given by : E.ds.

since, the flux through the cube would be $E\times a2 = x$

therefore for a sphere,  the flux would be $E.\times Φ a2$

which is equal to $Φ$

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

The law that describes the force as directly proportional to magnitude of charges and inversely proportional to the distance between the charges is known as :

  1. Newton's law

  2. Coulomb's law

  3. Gauss's law

  4. Ohm's law
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Newton's law : force $F=ma$  where m=mass and a=acceleration

Coulomb's law : force $F=\dfrac{kq _1q _2}{r^2}$ where $q _1, q _2 $ are charges and $r=$ separation of charges and $k=$ proportionality constant.
Gauss's law : the electric flux $\phi=\dfrac{q}{\epsilon _0}$ 
Ohm's law : Potential across a wire of resistance R is $V=IR$ where I is the current. 

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

When we wear nylon dresses during winter then there is ______ current which gets produced due to contact with out body. Fill in the Blank.

  1. Magnetic

  2. Electrostatic

  3. Potential

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

When we wear nylon dresses during winter then there is electrostatic current which gets produced due to contact with out body.


Option B is correct.

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

 A point charge Q is placed at origin O. Let $\overrightarrow {{E _A}} $,$\overrightarrow {{E _B}} $ and $\overrightarrow {{E _C}} $ represent electric fields at A, B and C respectively. If coordination of A,B and C are respectively (1,2,3) m,(1,1,-1) m and (2,2,2) m  then 

  1. $\overrightarrow {{E _A}} \bot \overrightarrow {{E _B}} $
  2. $\overrightarrow {{E _A}} \parallel \overrightarrow {{E _B}} $
  3. $\left| {\overrightarrow {{E _B}} } \right|\parallel 4\left| {\overrightarrow {{E _C}} } \right|$
  4. $\left| {\overrightarrow {{E _B}} } \right|\parallel 8\left| {\overrightarrow {{E _C}} } \right|$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\overrightarrow{E _a}=\cfrac{h _E}{ra^3}=\cfrac{h _E}{1^2+2^2+3^2}(1\hat{i}+2\hat{j}+3\hat{k})\ \overrightarrow{E _a}=\cfrac{h _E}{14^{3/2}}(1\hat{i}+2\hat{j}+3\hat{k})\ \overrightarrow{E _b}=\cfrac{ha}{r _b^2}\overrightarrow{r _b}=\cfrac{h _E}{(1^2+1^2+1^2)}^{3/2}(1\hat{i}-1\hat{j}+1\hat{k})\=\cfrac{h _E}{3^{3/2}}(1\hat{i}-1\hat{j}+1\hat{k})\ \overrightarrow{E _c}=\cfrac{h _E}{r _c^2}\overrightarrow{r _c}=\cfrac{h _E}{(2^2+2^2+2^2)^{3/2}}(2\hat{i}+2\hat{j}+2\hat{k})$

$\quad=\cfrac{h _E}{12^{3/2}}(2\hat{i}+2\hat{j}+2\hat{k})$
Now 
$\overrightarrow{E _a}.\overrightarrow{E _b}=\cfrac{h _E}{14^{3/2}}(1\hat{i}+2\hat{j}+3\hat{k})=\cfrac{h _E}{3^{3/2}}(1\hat{i}-1\hat{j}+1\hat{k})\ \Rightarrow \overrightarrow{E _a}.\overrightarrow{E _b}=(\cfrac{h _E}{14^{3/2}})(\cfrac{h _E}{3^{3/2}})(1-2+3)\neq0$
Thus$\overrightarrow{E _a}$ and $\overrightarrow{E _b}$ are perpendicular to each other
$|E _c|=\cfrac{h _E}{12^{3/2}}(2^2+2^2+2^2)^{1/2}=\cfrac{h _E}{12^{3/2}}(12)^{1/2}\ \Rightarrow |\overrightarrow{E _c}|=\cfrac{h _E}{12}\ \overrightarrow|E _b|=\cfrac{ha}{3^{3/2}}(1^2+1^2+1^2)^{1/2}=\cfrac{ _E}{3}=4\times\cfrac{h _E}{12}=4|\overrightarrow{E _c}|$
So, $|\overrightarrow{E _b}|=4|E _c|$



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

Four charges $+Q,-Q,+Q$ and $-Q$ are situated at the corners of a square; in a sequence then at the centre of the square:

  1. $E=0,V=0$
  2. $E=0,V\neq 0$
  3. $E\neq 0,V=0$
  4. $E\neq 0,V\neq 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Four charges $+Q$, $-Q$, $+Q$, $-Q$.

In this context, $E\neq 0$ but $V=0$ because. $E$ is not cancel out to each other but $V$ is cancel out each other.

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

Two charges Q and -2Q are placed at some distance. the locus of points in the plane of the charges where the potential is zero will be

  1. Straight line

  2. Circle

  3. Parabola

  4. ellipse
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Two charges $Q$, $-2Q$
some distance
Potential $=0$
hence,
$\dfrac { 1 }{ 4\pi { \epsilon  } _{ 0 } } \times \dfrac { Q\times \left( -2Q \right)  }{ R } =0$
When we use it the parabolic condition then,
${ y }^{ 2 }=4ax\quad \longrightarrow \left( 1 \right) $
Now,
$\dfrac { 1 }{ 4\pi { \epsilon  } _{ 0 } } \times \dfrac { Q\times 2Q }{ R } =0$
$\dfrac { { 2Q }^{ 2 } }{ R } =0\quad \longrightarrow \left( 2 \right) $
Hence, equating $(1)$ and $(2)$ and we get, the system is getting parabola.
Multiple choice physics electric fields introduction to electrostatic force electric force charging and discharging

Force of attraction between two point charges $Q$ and $-Q$ separated by $d$ meter is $F _e$. When these charges are placed two identical sphere of radius $R=0.3\ d$ whose centries are $d$ meter apart the force of attraction between them is 

  1. Greater than $F _{e}$
  2. Equal to $F _{e}$
  3. Less than $F _{e}$

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

Force of attraction between two point charges $Q$ and $-Q$ separated by $d$.

Force ${ F } _{ e }$
radius $=R=3d$
That is also equal to ${ F } _{ e }$ because the distance is same hence force is also same.