PL The capacitors in the figure have values: d,-1.00 F, G-2.00 F, C-3.00 F All t

Question

PL The capacitors in the figure have values: d,-1.00 F, G-2.00 F, C-3.00 F All three capacitors are initially uncharged. The switch S on the figure is first connected at position A so that the battery of voltage V=12.0V can charge capacitor C, with a charge q a. (4) Calculate the charge q, on the capacitor C b. (3) Calculate the total energy U, of all three capacitors when the switch S is connected to point A. c. (4) Now, assume that the C is a parallel plate capacitor and immersed in a material with the dielectric constant x-3.00. Calculate the new charge eW on this capacitor. Now the switch S is in position B and the capacitor C, is without the dielectric material d. (9) Determine the new charges

Explanation / Answer

1)

q3 = C3*V

= 3*10^-6*12

= 3.6*10^-5 C

b)

Total energy = 0.5*C3*V^2

= 0.5*3*10^-6*(12)^2

= 2.16*10^-4 J

c)

q3,new = k*q3 = 3*3.6*10^-5

= 10.8*10^-5 C

= 1.08*10^-4 C

d)

Let the charge on C1 and C2 be q

So, conserving charge,

q3 = q3' + q1'

So, 3.6*10^-5 = q3' + q1'

Also, voltage across C3 = voltage across C1 + C2

So, q3'/3 = q1'/1 + q1'/2

So, q1' = 6.55*10^-6 C , q3' = 2.95*10^-5 C , q2' = 2.95*10^-5 C

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