JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Capacitance

In a parallel plate capacitor set up, plate area of capacitor is

2 m^{2}

and the plates are separated by 1 m . If the space between the plates are filled with a dielectric material of thickness 0.5 m and area

2 m^{2}

(See fig) the capacitance of the set- up will be -------

ε_{0}

.
(Dielectric constant of the material

= 3.2

) (Round off to the Nearest Integer)

Answer: 3

Solution

The slab covers the full plate area and fills half the gap, so the gap behaves as two capacitors in series, each of separation

\frac{d}{2} = 0.5 m

and area

2 m^{2}

C_{1} = \frac{K ε _{0} A}{d /2} = \frac{3.2 ε _{0} \times 2}{0.5} = 12.8 ε_{0}, C_{2} = \frac{ε _{0} A}{d /2} = \frac{ε _{0} \times 2}{0.5} = 4 ε_{0}

Combining them in series,

C = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = \frac{12.8 \times 4}{12.8 + 4} ε_{0} = \frac{51.2}{16.8} ε_{0} = 3.05 ε_{0}

Rounded to the nearest integer, the capacitance of the set-up is

3 ε_{0}

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Integer answer