Question

A parallel plate capacitor is charged to potential \( V = 300\,V \) and then disconnected from the battery. If the separation between the plates is doubled, what will be the new potential difference across the plates?

Hint:

If a charged capacitor is isolated and plate separation changes, the charge stays constant. Use \( V = \frac{Q}{C} \) to calculate the new potential.

Answer 1

Step 1: Use the formula for potential: \[ V = \frac{Q}{C} \] Since battery is disconnected, charge \(Q\) remains constant. 
Capacitance of parallel plate: \[ C = \frac{\varepsilon_0 A}{d} \Rightarrow C \propto \frac{1}{d} \] If \(d\) is doubled, then \(C\) becomes \(C/2\). Hence, \[ V' = \frac{Q}{C/2} = 2V \Rightarrow V' = 2 \cdot 300 = 600\,V \] Final Answer: \[ \boxed{V' = 600\,V} \]