Question

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

Answer 1

charge, q = ± 1.1 mC

Given:

Capacitance,
`C = 10.0\micro F = 10.0* 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* 10^(- 6)* 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