Question

A 12.5-µF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them.

(a) How much energy is stored in the capacitor before and after the dielectric is inserted?
(b) By how much did the energy change during the insertion? Did it increase or decrease?

Answer 1

a) 0.0036 J & 0.0135 J

b) 0.0099 J

Step-by-step explanation:

Given that

Capacitance of a capacitor, C(i) = 12.5*10^-6 F

Potential difference across the capacitor, V = 24 V

Dielectric constant, k = 3.75

Energy in a capacitor is given by the formula

U = ½ CV²

Now, applying this to our solution, we have.

a, the energy stored before the dielectric was inserted is 0.0036. After inserting the dielectric, we then take cognizance into it, essentially multiplying by the constant of the dielectric, we got 0.0135 J

b) Now, Change in energy is gotten by subtracting the two energies, and we have

ΔU = U - U(1)

ΔU = 0.0036 - 0.0135

ΔU = 00099 J

See attachment for calculation