Question

A parallel-plate capacitor filled with air carries a charge Q. The battery is disconnected, and a slab of material with dielectric constant k = 2 is inserted between the plates. Which of the following statements is correct? The voltage across the capacitor decreases by a factor of 2. The charge on the plates is doubled. The voltage across the capacitor is doubled. The charge on the plates decreases by a factor of 2. The electric field is doubled.

Answer 1

Answer:

The voltage across the capacitor decreases by a factor of 2

Explanation:

When we put a dielectric material on a capacitor, we generate a polarization on it that changes the capacitance of the capacitor. The dielectric constant is the ratio between the initial capacitance (Co) without the dielectric and the new capacitance (C) with the dielectric:

$k=\frac{C}{C_0}$

So, if k is equal to 2:

$2= \frac{C}{C_0}$

$C=2*C_0$

So, the capacitance is doubled with the dielectric

Because the battery is disconnected, the charge on the plates of capacitor must be constant because conservation of charge.

The effective electric field of a capacitor is the electric filed when we put a dielectric on it, and it is:

$E_{eff}=\frac{E}{k}=\frac{E}{2}$

With E the initial electric field, so the electric field is halved

Finally, because we know electric field and voltage (V) on a parallel capacitor with distance between plates (d) are related by:

$E=\frac{V}{d}$

Because electric field and voltage are directly proportional and d remains constant if Electric field is halved, then voltage is halved too. The voltage across the capacitor decreases by a factor of 2.