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1 vote
What capacitance is needed to store 3μC of charge at a voltage of 120V?

1 Answer

2 votes

Answer:

C = 0.025F

Step-by-step explanation:

Charge =q=3×
10^(-6)C

Voltage=V=120V

Q=CV

C=Q/V

=3×
10^(-6)/120

=1/40×
10^(-6)

C = 0.025F

C=2.5
10^(-4)C 0r 25 ×
10^(-3)F

User Thomas Aylott
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