Question

A parallel-plate capacitor of capacitance 5 µF is connected to a battery of emf 6 V. The separation between the plates is 2 mm. (a) Find the charge on the positive plate. (b) Find the electric field between the plates. (c) A dielectric slab of thickness 1 mm and dielectric constant 5 is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination. (d) How much charge has flown through the battery after the slab is inserted?

Answer 1

a) Charge on the positive plate = 30 uC

b) Electric field between the plates = 3X103 V/m

c)New capacitance = 8.33 uF

d)Charge flown through the battery = 20 uC

Explanation:

For the given capacitor,

$C=5 \mathrm{uF}$

$\mathrm{V}=6 \mathrm{V}$

$d=2 \mathrm{mm}=2 \times 10^{-3} \mathrm{m}$

(a)  The charge on the positive plate is calculated using

$q=5 \mathrm{uF} \times 6 \mathrm{V}=30 \mathrm{uC}$

(b) The electric field between the plates of the capacitor is given by

$E=\frac{V}{d}=3 \times 10^{3} \mathrm{V} / \mathrm{m}$

(c)

Separation between the plates of the capacitor, $d=2 \times 10^{-3} \mathrm{m}$

Dielectric Constant, k = 5

Thickness of Dielectric Constant, $t=1 \times 10^{-3} \mathrm{m}$

Area of the plates of the capacitor,

$C=\frac{\epsilon_{0} A}{d}$

     $=5 \times 10^{-6}\\\\=\frac{8.85 \times 10^{-12} \times A}{2 \times 10^{-3}}$

$10^{4}=8.85 \times A$

$A=10^{4} \times \frac{1}{8.85}$

When the dielectric is placed in it, the capacitance becomes

$C_{1}=\frac{\epsilon_{0} A}{d-t+\frac{t}{k}}$

$C_{1}=\frac{8.85 \times 10^{-12} \times \frac{10^{4}}{8.85}}{2 \times 10^{-3}-10^{-3}+\frac{10-3}{5}}$

$C_{1}=\frac{8.85 \times 10^{-12} \times \frac{10^{4}}{8.85}}{10^{-3}+\frac{10-3}{5}}$

$C_{1}=\frac{10^{-12} \times 10^{4} \times 5}{6 \times 10^{-3}}$

$=8.33 \mathrm{uF}$

(d) C = 5 X 10-6

    V = 6V

   Q = CV

       = 3 X 10-5

       = 30 uF

Charge flown = Q1-Q2

                        = 20 uC