Question

With the battery connected, fill the gap by a slab with the dielectric constant 3.4. If the potential is 30 V , the plate separation is 0.2 mm , and the plate area is 87.4 cm2 , find the electric charge on the plate. The value of the permittivity of free space is 8.8542 × 10−12 C 2 /N · m2 . Answer in units of C.

Answer 1

Q = 3.96 × 10⁻⁸ C

Step-by-step explanation:

given,

dielectric constant (K) = 3.4

potential = 30 V

plate separation = 0.2 mm

Plate area = 87.4 cm²

Capacitance =
`(K A \epsilon_0)/(d)`

`\C =(3.4* 87.4 * 10^(-4) * 8.8542 * 10^(-12))/(0.2* 10^(-3))`

C = 1.32 × 10⁻⁹ F

Q = C V

Q = 1.32 × 10⁻⁹ × 30

Q = 3.96 × 10⁻⁸ C