What charge appears on each plate of a 10.0 μF parallel plate capacitor, when it is charged to 110 V?
1 answer:
Answer:
charge, q = ± 1.1 mC
Given:
Capacitance,
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 =
Therefore, the charge on the positive plate is:
q = + 1.1 mC
the charge on the negative plate is:
q = - 1.1 mC
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