Question

evaluate the capacitance of 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 what will be the charges in each plate?

Answer 1

use c=epsilon*area/distance(use si units)
then charge=capacitance*voltage

Answer 2

Answer:

Charges in each plate is $Q=530.4\times 10^{-8}\ C$ .

Step-by-step explanation:

Formula

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

Where $E_{0}$ 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 × 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 ×  0.001 m

          = 0.002 m

V = 200 v

$E_{0} = 8.84\times 10^{-10}$

Putting all the value in the formula

$C =8.84\times 10^{-10}\times \frac{0.06}{0.002}$

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

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

Put the values in the Q = CV formula

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

$Q= 265.2\times 10^{-10}\times 2\times 100$

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

Now by using the exponent property

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

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

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

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