Answer:
charge, q = ± 1.1 mC
Given:
Capacitance, [tex]C = 10.0\micro F = 10.0\times 10^{- 6} F[/tex]
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 = [tex]10.0\times 10^{- 6}\times 110 = 1100\micro C = 1.1 mC[/tex]
Therefore, the charge on the positive plate is:
q = + 1.1 mC
the charge on the negative plate is:
q = - 1.1 mC
