Question

3. A parallel plate capacitor separated 10 cm by an air barrier is connected to a 100V battery. The capacitance of the capacitor is 1
picofarad while the battery is connected. Without disconnecting the battery, the parallel plates are moved so they are now 20 cm apart.
What happens to the energy stored in the capacitor?​

Answer 1

Answer:

Since the battery remains connected, V remains constant. C decreases as d increases (C ∝ 1/d) and UC = ½ CV2

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Answer 2

Answer:

The energy stored in the capacitor decreases by half, when the distance between the plates is changed from 10 cm to 20 cm.

Explanation:

Given the distance between the plates of a parallel plate capacitor is 10 cm.

Energy stored in the capacitor is given by

$U=\dfrac{1}{2} CV^{2}$

The capacitance of a parallel-plate capacitor of surface area A separated by a distance d is given by

$C=\dfrac{\varepsilon_{0} A}{d}$

where $\varepsilon_{0}$ is the electric permittivity of free space.

Using this, the energy stored in the capacitor is given by

$U=\dfrac{1}{2} \dfrac{\varepsilon_{0} A}{d}V^{2}$

In the given problem, all the other quantities remain same, only d is varied, therefore $U\propto \dfrac{1}{d}$

For 10cm distance between the plates, the energy stored in the capacitor is $U_{10}= \dfrac{1}{10}$

Now when the distance is increased to 20cm, the energy stored in the capacitor is  $U_{20}= \dfrac{1}{20}$

The change in the energy stored is

$\dfrac{U_{20}}{U_{10}} =\dfrac{\dfrac{1}{20}}{\dfrac{1}{10}} =\dfrac{1}{20}}\times {\dfrac{10}{1}} =\dfrac{1}{2}$

$\implies U_{20}} =\dfrac{1}{2}{U_{10}}$

Therefore, the energy stored in the capacitor decreases by half, when the distance between the plates is changed from 10 cm to 20 cm.