Question

What charge accumulates on the plates of a 2.0-μF air-filled capacitor when it is charged until the potential difference across its plates is 100 V?

Answer 1

Answer:

0.0002 C.

Explanation:

Charge: This can be defined as the ratio of current to time flowing in a circuit. The S.I unit of charge is Coulombs (C)

Mathematically, charge can be expressed as

Q = CV ................................. Equation 1

Where Q = amount of charge, C = capacitance of the capacitor, V = potential difference across the plates.

Given: C = 2.0-μF = 2×10⁻⁶ F, V = 100 V.

Substitute into equation 1

Q = 2×10⁻⁶× 100

Q = 2×10⁻⁴ C

Q = 0.0002 C.

The amount of charge accumulated = 0.0002 C

Answer 2

Answer:

The capacitor accumulates 0.2 mC

Explanation:

The capacitance (with units Faraday) of a capacitor with a charge Q (with units Columbus) and a potential difference ΔV (with units Volts) is:

$C=\frac{Q}{\varDelta V}$ (1)

In our case C=$2.0\times10^{-6}F$ and ΔV=$100V$, so we can solve (1) for Q and use those values:

$C=\frac{Q}{\varDelta V}$

$Q=C{\varDelta V}=(2.0\times10^{-6}F)(100V)$

$Q=2.0\times10^{-4}C$