Question

A charge of 4.9 x 10-11 C is to be stored on each plate of a parallel-plate capacitor having an area of 150 mm2 and a plate separation of 3.3 mm. Also, the value of ε0 is 8.85 x 10-12 F/m. (a) What voltage (in V) is required if a material having a dielectric constant of 6.3 is positioned within the plates?

Answer 1

The voltage required when a material with a dielectric constant of 6.3 is positioned within the plates of the parallel-plate capacitor is approximately 96.1 V.

Step-by-step explanation:

First, we can calculate the capacitance of the parallel-plate capacitor using the formula:

C = (ε0 x A) / d

where C is the capacitance, ε0 is the vacuum permittivity (ε0 = 8.85 x 10^-12 F/m), A is the area of the plates (A = 150 mm^2 = 150 x 10^-6 m^2), and d is the separation between the plates (d = 3.3 mm = 3.3 x 10^-3 m).

By substituting the given values into the formula, we get:

C = (8.85 x 10^-12 F/m x 150 x 10^-6 m^2) / (3.3 x 10^-3 m) = 5.09 x 10^-13 F

Next, we can use the equation:

V = Q / C

to calculate the voltage, where V is the voltage, Q is the charge, and C is the capacitance.

By substituting the given charge (Q = 4.9 x 10^-11 C) and the calculated capacitance (C = 5.09 x 10^-13 F), we get:

V = (4.9 x 10^-11 C) / (5.09 x 10^-13 F) = 96.1 V

Therefore, the voltage required when a material with a dielectric constant of 6.3 is positioned within the plates is approximately 96.1 V.