Question

A certain parallel plate capacitor with capacitance 12.0μF is connected to a source of EMF with potential 3.0 V. A dielectric material with K=4.0 is then inserted between the plates of the capacitor. By how much does the energy stored in the capacitor change?

Answer 1

The change in energy stored in the capacitor when a dielectric material is inserted between the plates can be calculated using the formulas for energy and capacitance. In this case, the change in energy is 72.0μJ.

Step-by-step explanation:

The energy stored in a capacitor can be calculated using the formula:

E = (1/2) * C * V^2

Where E is the energy, C is the capacitance, and V is the potential difference.

In this case, the initial energy stored in the capacitor can be calculated as:

E_initial = (1/2) * 12.0μF * (3.0V)^2

Once the dielectric material is inserted, the capacitance changes. The new capacitance can be calculated using the formula:

C_new = C_initial * K

Where C_new is the new capacitance, C_initial is the initial capacitance, and K is the dielectric constant. Substituting the given values, we have:

C_new = 12.0μF * 4.0 = 48.0μF

Using the new capacitance, we can calculate the final energy stored in the capacitor as:

E_final = (1/2) * 48.0μF * (3.0V)^2

The change in energy can be calculated as:

ΔE = E_final - E_initial

Substituting the values, we get:

ΔE = (1/2) * 48.0μF * (3.0V)^2 - (1/2) * 12.0μF * (3.0V)^2

Simplifying, we find that the change in energy is 72.0μJ.