A parallel plate capacitor has a charge on one plate of q = 4.5E-07 C. Each square plate is d1 = 1.7 cm wide and the plates of the capacitor are separated by d2 = 0.35 mm. The gap is filled with air, εo = 8.85 × 10-12 C2/Nm2.

Part (a) What is the voltage between the plates, Δ V, in V?

Part (b) What plate width would double this voltage, in centimeters?

Explanation:

capacitance of capacitor

= ε₀ A / d , A is area of plate , d is plate separation

= (8.85 x10⁻¹² x 1.7² x 10⁻⁴) /( .35 x 10⁻³ )

= 73 x 10⁻¹³ F

Charge = 4.5 x 10⁻⁷

voltage between the plates,

Δ V = charge / capacitance

= 4.5 x 10⁻⁷ / 73 x 10⁻¹³

= .0616 x 10⁶

= 616 x 10² V

To make the vottage double , we will reduce capacitance to half .

To reduce capacitance half we will increase plate distance to double.

new plate distance

= 1.7 x 2

= 3.4 cm