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Question: Keeping the voltage of the charging source constant, what would be the percentage change in the ener...

Keeping the voltage of the charging source constant, what would be the percentage change in the energy stored in a parallel plate capacitor if the separation between its plates were to be decreased by 10%?

Explanation

Solution

The energy which is stored in a capacitor may be defined as the function of the voltage across the capacitor and the value of capacitance. And the capacitance C is inversely proportional to the distance between the parallel plates.
Also given that the separation distance between two plates is decreased by 10%. Hence calculate the capacitance in the two cases. That is, in the initial case and when the distance between the plates is decreased by 10%. By taking their ratio we will get the new capacitance. Then by substituting the value of initial and new capacitance we will get the change stored in a parallel plate capacitor.

Complete step-by-step solution: -
Energy can be calculated using the equation,
E=12CV2E=\dfrac{1}{2}C{{V}^{2}}……………..(1)
Then the capacitance C is given by,
C=ε0AdC=\dfrac{{{\varepsilon }_{0}}A}{d}………………….(2)
From this equation it is clear that capacitance C is inversely proportional to the distance between the parallel plates.
Since the separation distance between two plates is decreased by 10%. Then,
C1=ε0Ad1{{C}_{1}}=\dfrac{{{\varepsilon }_{_{0}}}A}{{{d}_{1}}}………………….(3)
d1=d0.1d d1=0.9d \begin{aligned} & {{d}_{1}}=d-0.1d \\\ & {{d}_{1}}=0.9d \\\ \end{aligned}
Dividing equation (3) by equation (2) we get,
C1C=dd1\dfrac{{{C}_{1}}}{C}=\dfrac{d}{{{d}_{1}}}
C1C=d0.9d\dfrac{{{C}_{1}}}{C}=\dfrac{d}{0.9d}
C1=1.11C\therefore {{C}_{1}}=1.11C
Then the change in energy is,
E1E=12×CV2(1.111){{E}_{1}}-E=\dfrac{1}{2}\times C{{V}^{2}}\left( 1.11-1 \right)
E1E=12×CV2(0.11){{E}_{1}}-E=\dfrac{1}{2}\times C{{V}^{2}}\left( 0.11 \right)
E1E=(0.11)×E{{E}_{1}}-E=\left( 0.11 \right)\times E
Thus the percentage change in energy is 11%.

Note: The energy which is stored in a capacitor may be defined as the function of the voltage across the capacitor. The voltage will be higher when it is in a parallel connection so we can say that in parallel connection it can store more energy than in a series connection.