Name: Energy Density in a Capacitor
Uploaded: 2019-07-19
Duration: 3 min 1 s
Description: The energy in the capacitor is stored in the form of the electric field. Let us derive a mathematical formula to calculate this energy in terms of the electr...

Question

A parallel plate capacitor of capacitance $100 μ F$ is charged to $200 V$ . After disconnecting it from the battery, using an insulated handle, the distance between the plates is doubled. Find:

$(i)$ potential difference between the plates and

(ii)

energy stored in the capacitor, after separation between the plates.

Accepted Answer

Let the parallel plate capacitor have an area $A$ and the distance between the plates be $d$ . The voltage across the capacitor is given to be $200 V$ . The capacitance is given as:

$(i)$

The charge stored on the capacitor is

$Q = C V = (100 \times 200) μ C = 0.02 C$

After increasing the distance between the plates, the capacitance will be:

$C' = \frac{ε_{0} A}{2 d} = \frac{C}{2}$

Since the battery was disconnected before increasing the distance between the plates, the charge on the capacitor will remain unchanged. Hence,

$Q = C' V' \Rightarrow V' = \frac{Q}{C'} = \frac{0.02}{\frac{C}{2}} = \frac{0.02 \times 2}{100 \times 10^{- 6}} = 400 V$

After increasing the distance between the plates, the capacitance will be:

$C' = \frac{ε_{0} A}{2 d} = \frac{C}{2}$

Potential difference between the plates of the capacitor, after increasing the plates distance is:

$V' = 400 V$

Hence, the energy stored in the capacitor after increasing the distance will be:

$U = \frac{1}{2} C' {(V')}^{2} \Rightarrow U = \frac{1}{2} (50 \times 10^{- 6}) \times {(400)}^{2} \Rightarrow U = 4 J$

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