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11Alexandr11 [23.1K]
4 years ago
8

A power source of 2.0 V is attached to the ends of a capacitor. The capacitance is 4.0 μF.

Physics
2 answers:
N76 [4]4 years ago
8 0
Voltage across capacitor, V = 2.0V
Capacitance of Capacitor, C = <span>4.0 μF
Charge stored in Capacitor, Q = ?

</span><span>The capacitance formula, C = Q/V
</span><span>We can rewrite this as Q = CV

Therefore, </span>the amount of charge stored in this capacitor = 2.0 * 4.0 <span>μC
= 8.0 </span>μC
Aleksandr-060686 [28]4 years ago
7 0
Answer:
Q = 8 μC

Explanation:
The relation between voltage, capacitance and charge can be expressed using the following rule:
Q = C * V
where:
Q is the amount of charge that we want to calculate
C is the capacitance = 4 * 10⁻⁶ F
V is the voltage applied = 2 V

Substitute with the givens in the above equation to get the amount of charge as follows:
Q = C * V
Q= 4 * 10⁻⁶ * 2
Q = 8 * 10⁻⁶ Coulumb
Q = 8 μC

Hope this helps :)
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