Question

What charge appears on each plate of a 10.0 μF parallel plate capacitor, when it is charged to 110 V?

Answer 1

Answer:

charge, q = ± 1.1 mC

Given:

Capacitance, $C = 10.0\micro F = 10.0\times 10^{- 6} F$

Voltage, V = 110 V

Solution:

The charge on the capacitor plates can be calculated by using the definition of capacitance as :

q ∝ V

where

q = charge

V = potential difference or Voltage

Therefore,

q = CV

Now, charge, q :

q = $10.0\times 10^{- 6}\times 110 = 1100\micro C = 1.1 mC$

Therefore, the charge on the positive plate is:

q = + 1.1 mC

the charge on the negative plate is:

q = - 1.1 mC