Question

Two 5.0 mm * 5.0 mm electrodes are held 0.10 mm apart and are attached to a 9.0 V battery. Without disconnecting the battery, a 0.10-mm-thick sheet of Mylar is inserted between the electrodes. What are the capacitor’s potential difference, electric field, and charge

(a) before and

(b) after the Mylar is inserted?

Short answer

(a) Before the Mylar is inserted, the capacitor’s potential difference, electric field, and the charge is$9V,9kW/m$and$20pC$respectively.

(b) After the Mylar is inserted, the capacitor’s potential difference, electric field, and the charge $9V,9kV/m$and$63pC$.

Step-by-step solution

Part (a) Step 1: Given information 

We need to find the capacitor’s potential difference, electric field, and charge before the Mylar is inserted.

Part (a) Step 2: Simplify

The potential difference will remain $9Volts$as the capacitor remains connected to the battery, before and after the insertion of the dielectric.

The electric field can be given as

$E=\frac{∆V}{d}=\frac{9}{0.0001}=9kV/m$,

The charge in the capacitor is given by

$q=CU=\in {\in }_0\frac{L^2}{d}U$,

where $\in$is the dielectric constant of the material. So, before insertion of the dielectric material, the charge is given as

$q=8.85·{10}^{-12}·\frac{0.{005}^2}{0.0001}·9=20pC$

Part (b) Step 1: Given Information

We need to find the electric field, and charge after the Mylar is inserted.

Part (b) Step 2: Simplify

The electric field will be constant even after the dielectric is inserted. It remains

$E=9kW/m$

And considering that after insertion, the Mylar the dielectric constant is $3.1$,then

$q_2=3.1·20=63pC$