What is the formula for capacitance of parallel plate capacitor?

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The formula for the capacitance of a parallel plate capacitor is C = ε0A/d

C = \(\epsilon_0A \over d\)

where C is the capacitance in farads (F)
ε₀ is the permittivity of free space (a constant equal to 8.85 x 10^-12 F/m)
A is the area of overlap between the two plates in square meters (m²)
d is the distance between the plates in meters (m)

Solved Example

Suppose we have a parallel plate capacitor consisting of two metal plates with an area of 0.1 m² each, and the distance between the plates is 0.001 m. We want to find the capacitance of this capacitor.

Solution: Using the formula for capacitance of a parallel plate capacitor, we have:

C = ε0A/d

where ε0 = 8.85 x 10^-12 F/m,
A = 0.1 m²,
d = 0.001 m.

Substituting these values into the formula, we get:

C = (8.85 x 10^-12 F/m) x 0.1 m²/0.001 m

Simplifying this expression, we get:

C = 8.85 x 10^-10 F

Therefore, the capacitance of the parallel plate capacitor is 8.85 x 10^-10 F.

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What is the formula for capacitance of parallel plate capacitor?

Solved Example

CBSE CLASS XII Related Questions

1.
Write any two features of nuclear forces.

2.
Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

3.
If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

4.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

5.
Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

6.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

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What is the formula for capacitance of parallel plate capacitor?

Solved Example

CBSE CLASS XII Related Questions

1. Write any two features of nuclear forces.

2. Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

3. If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

4.Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

5.Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

6.Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Write any two features of nuclear forces.

2.
Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

3.
If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

4.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

5.
Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

6.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.