Question

A parallel plate capacitor has two plates of area separated by a distance in air. the capacitor is connected to a 9-volt battery. You insert between the two plates a slab of plastic that has a dielectric constant k=6. What will happen to the capacitance of the capacitor and to the energy stored in the capacitor when you insert the dielectric?

Answer 1

Answer on question #77065 - Physics / Electric Circuits

A parallel plate capacitor has two plates of area separated by a distance in air. The capacitor is connected to a 9-volt battery. You insert between the two plates a slab of plastic that has a dielectric constant $k=6$. What will happen to the capacitance of the capacitor and to the energy stored in the capacitor when you insert the dielectric?

**Input Data:**

air permittivity: $\varepsilon_{air} = 1$;

plastic permittivity: $\varepsilon_{plastic} = 6$;

**Solution:**

The capacitance of the capacitor is found from the formula:

$$C = \varepsilon \frac{\varepsilon_0 * S}{d}$$

Energy stored in the capacitor:

$$W = \frac{C U^2}{2}$$

It is obvious that the capacitance is directly proportional to the dielectric constant, and the stored energy is directly proportional to the capacitance.

Then, substituting the values, we get:

$$C_{air} = 1 * \frac{\varepsilon_0 * S}{d};$$ $$C_{plastic} = 6 * \frac{\varepsilon_0 * S}{d}, \text{ then}$$ $$C_{plastic} = 6 * C_{air};$$ $$W_{air} = \frac{C_{air} U^2}{2};$$ $$W_{plastic} = \frac{6 * C_{air} U^2}{2};$$

**Answer:**

When you insert a plastic plate into the air gap of the capacitor, the power and stored energy will increase by a factor of 6. At the same time, there will be additional power consumption from the battery.

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