Question

evaluate the capacitance of a parallel plate capacitor having parallel plates of area 6 cm square placed at a separation of 2 mm considered area between plates as a dielectric medium if the capacitor is connected to 200 volt power supply what will be the charge on each plate

Answer 1

"We know that,

$C = \epsilon_0 \frac {A}{d}$

Where \epsilon is the permittivity of free space , A is the area of plate overlap d is distance between plates .

Q = CV

Where C is the capacitance and V is the voltage .

As given,

Parallel plate capacitor having parallel plates of area 6 cm square placed at separation of 2 mm consider air between plates as a dielectric medium of a capacitor is connected to 200 volt power supply.

As 1cm = 0.01m

Now convert 6cm into meter .

6 cm = $6 \times 0 .01$

        = 0.06 m

Area = 0.06 cm

d = 2 mm

As

1mm = 0.001m

Now convert 2 mm into m .

2 mm = $2 \times 0.001 m$

          = 0.002 m

V = 200 v

$\epsilon_0 = 8.84 \times 10^{-10}$

Putting all the value in the formula

$C = 8.84 \times 10^{-10} \times \frac {0.06}{0.02}$

$C = 8.84 \times 10^{-10} \times 30$

$C = 265.2 \times 10^{-10} farad$

Put the values in the Q = CV formula

$Q = 265.2 \times 10^{-10}\times 200$

$Q = 530.4 \times 10^{-10} \times {2}$

Now by using the exponent property

$x^a \times x^{-b} = x^{a-b}$

$Q = 530.4 \times 10^{-8} C$

Therefore the charges in each plate is $530.4 \times 10^{-8} C.$"