Question

A parallel-plate capacitor is connected to a battery. After it becomes charged, the capacitor is disconnected from the battery and the plate separation is increased. What happens to the potential difference between the plates? A parallel-plate capacitor is connected to a battery. After it becomes charged, the capacitor is disconnected from the battery and the plate separation is increased. What happens to the potential difference between the plates? The potential difference between the plates decreases. More information is needed to answer this question. The potential difference between the plates increases. The potential difference between the plates stays the same.

Answer 1

Answer:

The potential difference between the plates increases

Explanation:

As we know that the capacitance of the capacitor is given by:

$q = CV$         (1)

where

q = charge

C = capacitance

V = Voltage or Potential Difference

Also, the capacitance of a parallel plate capacitor is given as:

$C = \frac{\epsilon_{o}A}{D}$           (2)

where

$\epsilon_{o} = permittivity of free space or vacuum$

A = Area of the plates

D = Separation distance between the plates

Now, from eqn (1) and (2):

$V = \frac{qD}{A\epsilon_{o}}$

Now, from the above eqn  we can say that:

Potential difference depends directly on the separation distance between the plates of the capacitor and is inversely dependent on the area of the plates of the capacitor.

Therefore, after disconnecting, if the separation between the plates is increased the potential difference across it also increases.