What is the charge of the plates of a capacitor, the capacitance of which is 2 μF
| August 12, 2021 | education
What is the charge of the plates of a capacitor, the capacitance of which is 2 μF and the voltage between the plates is 20V?
C = 2 μF = 2 * 10 ^ -6 F.
U = 20 V.
Q -?
The capacitance of a capacitor C is the coefficient of proportionality between the voltage U and the charge Q: C = Q / U.
Q = C * U.
Q = 2 * 10 ^ -6 F * 20 V = 40 * 10 ^ -6 Cl.
Answer: the charge on the plates of the capacitor Q = 40 * 10 ^ -6 Cl.
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