Two 13.0-cm-diameter electrodes 0.59 cm apart form a parallel-plate capacitor. T

Question

Two 13.0-cm-diameter electrodes 0.59 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 14 Vbattery.  What is the charge on each electrode after insulating handles are used to pull the electrodes away from each other until they are 1.0 cm apart? What is the electric field strength inside the capacitor after insulating handles are used to pull the electrodes away from each other until they are 1.0 cm apart? . What is the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 1.0 cm apart? The electrodes remain connected to the battery during all of these processes.

Explanation / Answer

a) q1 = C*v
where C = o*A/d = 8.854x10^-12**r^2/d = 8.854x10^-12**(0.065)^2/0.0059 = 1.991884224 x 10^-11 F

So q1 = 1.991884224 x 10^-11*14 = 2.788637914 x 10^-10C and q2 = -2.788637914 x 10^-10C

b) E = /o = 2.788637914 x 10^-10/(8.854x10^-12**r^2) = 2.788637914 x 10^-10/(8.854x10^-12**(0.065)^2)
= 2372.881356 V/m

c) V = E*d = 2372.881356 V/m*0.0059 = 14.0V

d) Now C decreases to
C = o*A/d = 8.854x10^-12**r^2/d = 8.854x10^-12**(0.065)^2/0.01 = 1.175211692 x 10^-11 F

q1 = 1.991884224 x 10^-11*14 = 2.788637914 x 10^-10C and q2 = -2.788637914 x 10^-10C

Charge remains constant in this situation

e) E = 2372.881356 V/m

It remains the same

f) V = E*d =2372.881356 V/m * 0.01 = 23.72881356 V

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