A parallel plate capacitor with plate area 100.0cm 2 and air-gap separation .500

Question

A parallel plate capacitor with plate area 100.0cm2 and air-gap separation .500mm is connected to a 12.0 V battery, and fully charged. The batter is then disconnected.

A.) The plates are now pulled to a separation of 1.00mm. What is the magnitude of the charge on each capacitor plate and the magnitude of the E-field between the plates now? ( ?0 = 8.85 x 10-12 C2/Nm2)

B.) What is the potential difference across the plates now?

C.) How much work was required to pull the plates to their new separation?

D.) If a proton was released from the + plate and hit the - plate, what speed would it have?

Thanks :D

Explanation / Answer

(a) the capacitance of capacitor,
C = ?_0A/d
     = 8.85*10^-12*3.0*10^-4 /0.50*10^-3
     = 5.31*10^-12 F
the charge on the capacitor, q = CV
   q = 5.31*10^-12*12
       = 63.72 *10^-12 C

(b) if the plates are now pulled to a separation of 1.00 mm then the charge ,
   q = 63.72 *10^-12 C (the charge is remains same)

(c) Now the capacitance of the plates,
   C' = ?_0A/d
= 8.85*10^-12*3.0*10^-4 /1.0*10^-3
       = 2.655*10^-12 F
the new potential difference V' = CV/C' = 5.31*10^-12*12/2.655*10^-12
                                             V' = 24 V

(d) the work done W = V'q = 24*63.72 *10^-12
                              W = 1529.28*10^-12 J

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