Question

A parallel plate capacitor is made of two circular plates separated by a distance 5mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3xx10^4V//m the charge density of the positive plate will be close to:

Answer 1

Answer:

2.64 × 10⁹

Explanation:

as E = V ×d

v = E/d

V = 3× 10^4 / 5 ×10^-3

V = 6×10^6 volts

and

charge density = charge per unit area on plate = Q/A

as Q = CV

C = AE/d (here E = dielectric constant)

then Q= EV /d

= 6×10⁶ ×2.2 /5×10^-3

=2.64 ×10⁹