A parallel plate capacitor has a capacitance of 6.78 μF and it is connected to a 12V battery. What is the charge on the capacito
r?
1 answer:
Answer:
8.136×10⁻⁵ J
Explanation:
Applying,
Q = Cv................ Equation 1
Where Q = Charge on the capacitor, v = voltage of the battery, C = capacitance of the capacitor.
From the question,
Given: C = 6.78μF = 6.78×10⁻⁶ F, v = 12 V
Substitute these values into equation 1
Q = (6.78×10⁻⁶ )(12)
Q = 8.136×10⁻⁵ J
Hence the charge on the capacitor is 8.136×10⁻⁵ J
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