Tag: parallel plate capacitor

Quizzes (0) Questions (30)

Questions Related to parallel plate capacitor

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

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 -

  1. 200 V

  2. 180 V

  3. 160 V

  4. 140 V
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The decay follows V = V0 * e^(-t/RC). At t=1, 240 = 320 * e^(-1/RC), so e^(-1/RC) = 240/320 = 0.75. At t=2, V = 320 * e^(-2/RC) = 320 * (e^(-1/RC))^2 = 320 * (0.75)^2 = 320 * 0.5625 = 180 V.

Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

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

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

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

Two identical capacitors charged to potential V each store energy (1/2)CV^2, making the initial total energy CV^2. When connected with reversed polarity, the charges neutralize each other completely, resulting in a final potential and total energy of zero. The loss in electrostatic energy is therefore equal to the initial total energy, which is CV^2.

Multiple choice parallel plate capacitor electrostatic potential and capacitance electrostatics physics

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 :

  1. $=$T
  2. $>$T
  3. $<$T

  4. any of the above three is possible
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $T' < T$

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

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

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.