Question

A 2.45 nF parallel-plate capacitor is charged to an initial potential difference of 56 V and then isolated. The dielectric material between the plates has a dielectric constant of 5.6. How much work is required to withdraw the dielectric sheet? Answer in units of J

Answer 1

The work required to withdraw the dielectric sheet = 3.84×10⁻⁶ J

Step-by-step explanation:

Energy Stored in a capacitor = 1/2CV²

E =1/2CV²............................ Equation 1

Where E = Energy stored in the capacitor, C = capacitance of the capacitor, V = potential difference.

Note: The work required to withdraw the dielectric sheet is equal to the energy stored in the capacitor

Given: C = 2.45 nF = 2.45 × 10⁻⁹ F, V = 56 V

Substituting these values into equation 1

E = 1/2(2.45×10⁻⁹)(56)²

E = 1/2(2.45×10⁻⁹)(3136)

E = 1568(2.45×10⁻⁹)

E = 3841.6×10⁻⁹

E = 3.84×10⁻⁶ J.

Thus the work required to withdraw the dielectric sheet = 3.84×10⁻⁶ J