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

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

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