Question

An empty parallel plate capacitor is connected between the terminals of a 18.8-V battery and charges up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor

Answer 1

Answer:

$p.d' = 37.6 V$

Explanation:

From the question we are told that:

Potential difference $p.d=18.8V$

New Capacitor $C_1=C_2/2$

Generally the equation for Capacitor capacitance is mathematically given by

$C=\frac{eA}{d}$

Generally the equation for New p.d' is mathematically given by

 $C_2V=C_1*p.d'$

  $p.d' = 2V$

 $p.d'= 2 * 18.8$

 $p.d' = 37.6 V$