Tag: electrostatics

Questions Related to electrostatics

A glass slab is put within the plates of a charged parallel plate condenser. Which of the following quantities does not change?

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. energy of the condenser

  2. capacity

  3. intensity of electric field

  4. charge

Show answer
Correct Option: B

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 $9$.

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. $10^{7}V$.

  2. $10^{4}V$.

  3. $10^{5}V$.

  4. $10^{6}V$.

Show answer
Correct Option: B

A capacitor of capacitance 2 $\mu $ F is charged to a potential difference of 12 V. It is then connected across inductor of inductance 0.6 mH. The current in the circuit when the potential difference across the capacitor 6 V is :

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. 3.6 A

  2. 2.4 A

  3. 1.2 A

  4. 0.06 A

Show answer
Correct Option: D

The parallel plates of capacitor are charged to a potential difference of 320 volts and are then connected across a resistor. The potential difference across the capacitor decays exponentially with time. Alter 1 second the potential difference between the plates of the capacitor is 240 volts then after 2 seconds the potential difference between the plates will be -

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. 200 V

  2. 180 V

  3. 160 V

  4. 140 V

Show answer
Correct Option: B

When charge is supplied in a system where the capacitor are connnected in parallel. the charge will be distributed

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. Equally to all the capacitor

  2. indirect ratio to the capacitance value

  3. in an inverse ratio to the capacitance values

  4. both (b) and (c)

Show answer
Correct Option: C

Two identical capacitors are connected in parallel across a potenial difference V. after they are fully charged, the positive plate of first capacitor is connected to negative plate of second and negative plate of first is connected to positive plate of other. The loss of energy will be

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. $\dfrac { 1 }{ 2 } { CV }^{ 2 }$

  2. ${ CV }^{ 2 }$

  3. $\dfrac { 1 }{ 4 } { CV }^{ 2 }$

  4. Zero

Show answer
Correct Option: C

A simple pendulum of mass m charged negatively to q coulomb oscillates with a time period T in a downward electric field E such that mg > qE. If the electric field is withdrawn, the new time period :

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. $=$T

  2. $>$T

  3. $<$T

  4. any of the above three is possible

Show answer
Correct Option: C
Explanation:

Time period in the absence of electric field $T' = 2 \pi \displaystyle \sqrt{\frac{l}{g}}$

In the presence of electric field
$g _{eff}=(mg - qE) / m$
Therefore $T = 2 \pi \displaystyle \sqrt{\frac{l}{(mg - qE) / m}}$
$\displaystyle T'  = 2\pi \sqrt{\frac{l}{g}}$
or $T' < T$

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 :

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  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

Show answer
Correct Option: C

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

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. 8

  2. 6

  3. 0.1

  4. 3

Show answer
Correct Option: D
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, $\displaystyle F _g=F _e \Rightarrow mg=(ne)\frac{V}{d}$


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

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

Consider two bodies A and B of same capacitance.  If charge of -10C flows from body A to body B, then

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. the potential of body A increases.

  2. the potential of body B decreases

  3. the magnitude of change in potential in both bodies is same.

  4. All the above

Show answer
Correct Option: D
Explanation:

As we know $ V = \dfrac{Q}{C}$ , where letters have their respective meanings.

So, if charge of $-10C$  (Negative Charge) flows from body A to body B, then the potential of body A increases and the potential of body B decreases. Also the magnitude of change in potential in both bodies is same as A and B are of same capacitance.
Therefore, D is correct option.