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?

the energy is half of the initial energy.

Explanation:

After the parallel plates capacitor becomes fully charged and it is disconnected from the battery and then the distance between the capacitor plates is is doubled such that there is no leakage of charge.

Energy of a capacitor is given as:

...................(1)

where:

C = capacitance of the capacitor

V = potential difference between the capacitor plates

And we also have:

....................(2)

where:

A = area of the capacitor plate

d = distance between the capacitor plate

permittivity of the medium between the plates

When the distance between the plates is doubled then the capacitance decreases to half:

Now form eq. (1) the energy of capacitor after the change:

the energy is half of the initial energy.

when a device is used in a circuit in which the voltage is 81 v, the current flowing through the device is 3 a. what is the resistance of the device?

a. 27 ω

b. 40.5 a

c. 243 ω

d. 243 a

the correct answer is a. 27