Question

Calculate the capacitance of a parallel plate capacitor, when a dielectric slab having a

dielectric constant x and thickness doubled the separation of the plates inserted between the (1 Mark)

plates.​

Answer 1

Answer:

Given:

Charge on the positive plane = +Q1

Charge on the negative plate = -Q2

To calculate: Charge on the capacitor

Let ABCD be the Gaussian surface such that faces AD and BC lie inside plates X and Y, respectively.

Let q be the charge appearing on surface II. Then, the distribution of the charges on faces I, III and IV will be in accordance with the figure.

Let the area of the plates be A and the permittivity of the free space be∈0.

Now, to determine q in terms of Q1 and Q2, we need to apply Gauss's law to calculate the electric field due to all four faces of the capacitor at point P. Also, we know that the electric field inside a capacitor is zero.

Electric field due to face I at point P, E1 =

Q1-q

2∈0A

Electric field due to face II at point P, E2 =

+q

2∈0A

Electric field due to face III at point P, E3 =

-q

2∈0A

Electric field due to face IV at point P, E4 = -(

-Q2+q

2∈0A

) (Negative sign is used as point P lies on the LHS of face IV.)

Since point P lies inside the conductor,

E1 + E2 + E3 + E4 = 0

∴ Q1 - q + q - q - (-Q2 + q ) = 0

⇒q =

Q1+Q2

2

Thus, the charge on the capacitor is

Q1+Q2

2

, which is the charge on faces II and III.