Question

A 4.00 µf capacitor is connected to a 12.0 v battery.

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 capacitance of the capacitor is:
$C=4 \mu F=4 \cdot 10^{-6}F$
and the voltage applied across its plates is
$V=12.0 V$

The relationship between the charge Q on each plate of the capacitor, the capacitance and the voltage is:
$C= \frac{Q}{V}$
and re-arranging it we find the charge stored in the capacitor:
$Q=CV=(4 \cdot 10^{-6} F)(12.0 V)=4.8 \cdot 10^{-5} C$

(b) The electrical potential energy stored in a capacitor is given by
$U= \frac{1}{2}CV^2$
where C is the capacitance and V is the voltage. The new voltage is 
$V=1.50 V$
so the energy stored in the capacitor is
$U= \frac{1}{2}(4 \cdot 10^{-6} F)(1.50 V)^2=4.5 \cdot 10^{-6} J$