Answer:
q' = 0.5q
Hence, the charge stored is also halved
Explanation:
The capacitance of a capacitor is given by the following formula:
[tex]C = \frac{q}{V}[/tex] ------- eqn(1)
where,
C = Capacitance of the capacitor
q = charge stored in the capacitor
V = potential across the capacitor
Now, if we make potential half of its initial value:
[tex]C = \frac{q'}{0.5V}\\\\CV = 2q'\\using\ eqn(1):\\\frac{q}{V}(V) = 2q'\\\\[/tex]
q' = 0.5q
Hence, the charge stored is also halved
