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There are six identical metal plates arranged as shown in the figure. The area of each plate is A and the separation between two consecutive plates is d. The total energy stored by this system when connected to a cell of emf V is:

Option: 1

\frac{\epsilon_{0}A}{d} v^2


Option: 2

\frac{1}{2} \frac{\epsilon_{0}AV}{d}


Option: 3

\frac{1}{2}\frac{\epsilon_{0}Av^2}{d}


Option: 4

None


Answers (1)

best_answer

The effective capacitance is:

                C = \frac{\epsilon_{0}A}{d}

So energy stored is:

                E = \frac{1}{2}CV^2

               E = \frac{1}{2}\frac{\epsilon_{0}A}{d} v^2

where, 

                C = \frac{\epsilon_{0}A}{d} (capacitance of parallel plate capacitor)

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Gunjita

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