Question

A parallel-plate capacitor connected to a battery becomes fully charged. After the capacitor from the battery is disconnected, the separation between the plates of the capacitor is doubled in such a way that no charge leaks off. 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 remains constant. C.)The energy stored in the capacitor doubles its original value. D.)The energy stored in the capacitor is decreased to one-fourth of its original value. E.)The energy stored in the capacitor is decreased to one-half of its original valu

Answer 1

Answer:

The correct answer is D.

Explanation:

Parallel Plate Capacitors are made up of two plates that are connected to each other and there is a set distance between them which is filled with various materials such as air, mika or paper etc. which all affect the capacitance of the capacitor which is it's ability to store electrical charge.

According to the capacitance formula  $C=\frac{kEA}{d}$ where C is the capacitance, k is the value of permittivity of the material between the plates, A is the surface area of the plates, E is the constant value of permittivity of free space and d is the distance between the plates, if we double the value of d, the energy stored in the capacitor will be decreased to half of its original value.

I hope this answer helps.