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The voltage applied across a given parallel-plate capacitor is doubled. How is the energy stored in the capacitor affected?

a.The energy stored in the capacitor quadruples its original value.

b.The energy stored in the capacitor doubles its original value.

c.The energy stored in the capacitor is decreased to one-fourth of its original value.

d.The energy stored in the capacitor remains constant.

e.The energy stored in the capacitor is decreased to one-half of its original value.

Respuesta :

Answer:

(a) The energy stored in the capacitor quadruples its original value.

Explanation:

Consider C be the capacitance of the parallel plate capacitor and V be the voltage applied across the parallel plate capacitors.

Thus, the original energy stored in the capacitor is determine by the relation:

[tex]E_{1}=\frac{1}{2} CV^{2}[/tex]    ...(1)

Now, the voltage across the parallel plate capacitors becomes double while the capacitance remains same, that is,

New voltage = 2V

New Capacitance = C

Thus, the new energy stored in the capacitor is:

[tex]E_{2}=\frac{1}{2} C(2V)^{2}[/tex]

[tex]E_{2}=4\frac{1}{2} CV^{2}[/tex]

Substitute equation (1) in the above equation.

[tex]E_{2}=4E_{1}[/tex]

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