Question

A 4.00 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?

Answer 1

Explanation:

It is given that,

Capacitance of the capacitor, $C=4\ \mu F=4\times 10^{-6}\ F$

Voltage, V = 12 V

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

$C=\dfrac{Q}{V}$

$Q=C\times V$

$Q=4\times 10^{-6} \times 12$

$Q=4.8\times 10^{-5}\ C$

(b) Voltage, V = 1.5 V

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

$U=\dfrac{1}{2}CV^2$

$U=\dfrac{1}{2}\times 4\times 10^{-6}\times (1.5)^2$

$U=4.5\times 10^{-6}\ V$

Hence, this is the required solution.

Answer 2

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

A. Determination of the charge

•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

B. Determination of the energy stored in the capacitor.

•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

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