A4.00 µf capacitor is connected to a 12.0 v battery. a. what is the charge on each plate of the capacitor? b. if this same capacitor is connected to a 1.50 v battery, how much electrical potential energy is stored?

A. The charge on each plate of the capacitor is 4.8×10¯⁵ C

B. The energy stored stored on the capacitor is 4.5×10¯⁶ J

•Capacitor (C) = 4 µF = 4×10¯⁶

•Voltage (V) = 12 V

•Charge (Q) =?

Q = CV

Q = 4×10¯⁶ × 12

Q = 4.8×10¯⁵ C

•Voltage (V) = 1.5 V

•Capacitor (C) = 4 µF = 4×10¯⁶

•Energy (E) =?

E = ½CV²

E = ½ × 4×10¯⁶ × 1.5²

E = 2×10¯⁶ × 2.25

E = 4.5×10¯⁶ J

Learn more about energy stored in a capacitor:

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Explanation:

It is given that,

Capacitance of the capacitor,

Voltage, V = 12 V

(a) The charge stored per unit potential difference is called capacitance of the capacitor.

(b) Voltage, V = 1.5 V

The electrical potential energy stored in the capacitor is given by :

Hence, this is the required solution.