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17 June, 22:44

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?

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  1. 17 June, 23:00
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    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
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