Question

Design a parallel plate capacitor, including a dielectric, with a-capacitance of 150 nF. (Choose all physical parameters and demonstrate that they produce this capacitance.) How much voltage is required to hold 200nC of charge on each plate?

Answer 1

To design a parallel plate capacitor with a capacitance of 150 nF using polystyrene as the dielectric and a plate separation of 0.5 mm, the plate area should be approximately 104 cm². To hold 200 nC of charge on each plate, a voltage of 1.33 V is required.

Step-by-step explanation:

To design a parallel plate capacitor with a capacitance of 150 nF, we must select the appropriate physical dimensions and materials for the capacitor plates and dielectric. The capacitance (C) of a parallel plate capacitor is given by the formula:

C = ε0 εr (A/d)

where ε0 is the vacuum permittivity (8.854 x 10-12 F/m), εr is the relative permittivity of the dielectric material, A is the area of the plates, and d is the distance between the plates.

Assuming we use a common dielectric like polystyrene (εr ≈ 2.56), we can rearrange the formula to solve for A:

A = C × d / (ε0 εr)

If we choose a plate separation of 0.5 mm (0.0005 m), we can find A:

A = 150 x 10-9 F × 0.0005 m / (8.854 x 10-12 F/m × 2.56) ≈ 0.0104 m2 or 104 cm2

Now, to store a charge (Q) of 200 nC on each plate, we use the formula:

V = Q/C

V = 200 x 10-9 C / 150 x 10-9 F = 1.33 V

Therefore, a voltage of approximately 1.33 V is required to hold 200 nC of charge on each plate of the capacitor.