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A power source of 2.0 V is attached to the ends of a capacitor. The capacitance is 4.0 μF. What is the amount of charge stored in this capacitor? 0.50 μC 2.0 μC 6.0 μC 8.0 μC
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Feb 20, 2019
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A power source of 2.0 V is attached to the ends of a capacitor. The capacitance is 4.0 μF.
What is the amount of charge stored in this capacitor?
0.50 μC
2.0 μC
6.0 μC
8.0 μC
Physics
middle-school
Ralfeus
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Voltage across capacitor, V = 2.0V
Capacitance of Capacitor, C =
4.0 μF
Charge stored in Capacitor, Q = ?
The capacitance formula, C = Q/V
We can rewrite this as Q = CV
Therefore,
the amount of charge stored in this capacitor = 2.0 * 4.0
μC
= 8.0
μC
Iakovos
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Feb 23, 2019
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Iakovos
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Answer:
Q = 8 μC
Step-by-step 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 :)
Rafael Korbas
answered
Feb 25, 2019
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Rafael Korbas
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