Question

a parallel plate capacitor has circular plates of radius 8cm and plate separation 1mm what will be the charge on the plates if a pdf of 100 v ​

Answer 1

Answer:

Radius of circular plate = 8cm

Separation between the plates = 1mm

Pottential difference applied = 100V

To Find :

The charge on the plates of the capacitor = ?

Solution :

The capacitance of the plates C = \frac{A\epsilon}{d}C=

d

Aϵ

C = \frac{A\epsilon}{d}C=

d

Aϵ

C = \frac{\pi r^{2} \epsilon}{d}C=

d

πr

2

ϵ

C = \frac{\pi \times 0.08^{2}\times 8.85\times 10^{-12} }{1\times10^{-3}}C=

1×10

−3

π×0.08

2

×8.85×10

−12

C=1.7\times 10^{-10} FC=1.7×10

−10

F

The capacitance of the capacitor is 1.7×10⁻¹⁰ F

The charge on the capacitor plates (Q) = C×V

Q= 1.7\times 10^{-10} \times 100Q=1.7×10

−10

×100

Q= 17 nCQ=17nC

The charge on the capacitor plates is 17nC

Answer 2

Answer:

Given :

Radius of circular plate = 8cm

Separation between the plates = 1mm

Pottential difference applied = 100V

To Find :

The charge on the plates of the capacitor = ?

Solution :

The capacitance of the plates C = \frac{A\epsilon}{d}

C = \frac{A\epsilon}{d}

C = \frac{\pi r^{2} \epsilon}{d}

C = \frac{\pi \times 0.08^{2}\times 8.85\times 10^{-12} }{1\times10^{-3}}

C=1.7\times 10^{-10} F

The capacitance of the capacitor is 1.7×10⁻¹⁰ F

The charge on the capacitor plates (Q) = C×V

Q= 1.7\times 10^{-10} \times 100

Q= 17 nC

The charge on the capacitor plates is 17nC

Explanation:

hope it helps u

mark as brainliest

follow me

physics expert