please help. I am having a hard time understanding this problem. A.) Two sheets

Question

please help. I am having a hard time understanding this problem.

A.) Two sheets of aluminum foil have the same area, a separation of 1.9 mm, and a capacitance of 1350 pF, and are charged to 12 V. Calculate the area of each sheet.

B.) The separation is now decreased by 0.10 mm with the charge held constant. What is the new capacitance?

C.) By how much does the potential difference change? (Give the difference as the new potential minus the one before the decrease in separation). Explain how a microphone might be constructed using this principle.

Thanks,
Ryan

Explanation / Answer

Two sheets of alumonium foil have the same area A seperation distance d = 1.9 *10^-3 m capacitance C = 1350 pF Q = CV        = 1350 *10^-12 F * 12 V        = 16.2 *10^-9 C Area of the sheet is C = _0 A / d                              1350 *10^-12 F = 8.85 *10^-12 * A / 1.9*10^-3 m                                     A = 0.28 m^2 b) seperation distance d = 0.10 *10^-3 m              new capacitance C   = _0 A / d                                              = 8.85 *10^-12 * 0.28 m^2 / 0.10 *10^-3 m                                              = 24.7 *10^-9 F            V_0 =Q / C                    = 16.2 *10^-9 C / 24.7 *10^-9 F                   = 0.655V                         c ) potential difference is V - V_0                                       12.0 V- 0.655 V                                    = 11.34 V        

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