Question 82: Electrostatics & Capacitance

Question

A parallel plate capacitor has charge $5 	imes 10^{-6} \mathrm{C}$. A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is $4 	imes 10^{-6} \mathrm{C}$ then the dielectric constant of the slab is _______________.

Accepted Answer

To find the dielectric constant $ K $ of the slab, we use the formula for the induced charge $ Q_{	ext{in}} $ in a parallel plate capacitor:
$ Q_{	ext{in}} = Q \left(1 - \frac{1}{K}ight) $
Given:
Total charge $ Q = 5 	imes 10^{-6} \, 	ext{C} $
Induced charge $ Q_{	ext{in}} = 4 	imes 10^{-6} \, 	ext{C} $
We substitute the values into the formula:
$ 4 	imes 10^{-6} = 5 	imes 10^{-6} \left(1 - \frac{1}{K}ight) $
To solve for the dielectric constant $ K $, simplify the equation:
Divide both sides by $ 5 	imes 10^{-6} $:
$ \frac{4}{5} = 1 - \frac{1}{K} $
Rearrange to solve for $ \frac{1}{K} $:
$ 1 - \frac{4}{5} = \frac{1}{K} $
$ \frac{1}{5} = \frac{1}{K} $
Invert both sides to find $ K $:
$ K = 5 $
Therefore, the dielectric constant of the slab is $ K = 5 $.

Electrostatics & Capacitance

A parallel plate capacitor has charge 5 \times 10^{-6} \mathrm{C}. A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is 4 \times 10^{-6} \mathrm{C} then the dielectric constant of the slab is _______________.

Integer / Numeric Answer Type