Consider a simple parallel-plate capacitor whose plates are given equal and opposite charges and are separated by a distance d. Suppose the plates are pulled apart until they are separated by a distance D > d. The electrostatic energy stored in the capacitor is:

a. Greater than
b. The same as
c. Smaller than

c. Smaller than.

Explanation:

The energy stored in a capacitor is given as

E = 1/2CV² Equation 1

Where E = Energy stored in a capacitor, C = capacitance of the capacitor, V = Voltage.

Also,

C = εA/d Equation 2

Where ε = permitivity of the material, A = cross-sectional area of the plate, d = distance of separation of the plate.

substitute equation 2 into equation 1

E = 1/2εAV²/d Equation 3

From equation 3 above,

The energy stored in a capacitor is inversely proportional to the distance of separation between the plates.

Hence when the plate is pulled apart by a distance D (D>d) The energy stored in the capacitor will be smaller.

The right option is c. Smaller than.

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j