Two metal electrodes of size 5.0 cm \\times 5.0 cm are spaced 1.2 mm apart and c

Question

Two metal electrodes of size 5.0 cm imes 5.0 cm are spaced 1.2 mm apart and connected by wires to the terminals of a 9.0V battery.

A) What is the charge on each electrode?
Express your answer using two significant figures. (pC)

B) What is the potential difference between electrodes?
Express your answer using two significant figures. (V)

C) The wires are disconnected, and insulated handles are used to pull the plates apart to a new spacing of 2.4 mm. What is the charge on each electrode?
Express your answer using two significant figures. (pC)

D) What is the potential difference between electrodes?
Express your answer using two significant figures. (V)

Explanation / Answer

Given that Area of the metal plates A = 5.0 cm x 5.0 cm                                           = 25.0*10-4 m2, Seperation between the plates d = 1.2 mm, Potential of the battery, V = 9.0 V

(a) The capacitance of the parallel plate capacitor                                      C = 0A/d      Therefore the charge on each electrode                                      Q = CV                                          = 0AV/d                                          = (8.85*10-12 C2/Nm2)(25.0*10-4 m2)(9 V) /1.2*10-3 m                                          = 1.66*10-10 C                                          = 166 pC --------------------------------------------------------------------------------------------------------------------------- (b)       The potential difference between electrodes = 9.0 V --------------------------------------------------------------------------------------------------------------------------- (c) The new spacing d' = 2.4 mm = 2d
      Therefore the charge on each electrode is unchanged since they are not contact with battery, so        the charge remains same as                      Q' = Q = 166 pC
      The new capatitance C' = 0A/(2d) = C/2

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