Capacitor C_1 is connected alone to a battery and charged until the magnitude of

Question

Capacitor C_1 is connected alone to a battery and charged until the magnitude of the charge on each plate is 4.0 times 10^8 C. Then it is removed from the battery and connected to two other capacitors C_2 and C_3, as shown. The charge on the positive plate of C_1 is then 1.0 times 10^8 C. The charges on the positive plates of C_2 and C_3 are: A) q_2 = 2.0 times 10^-8 C and q_3 = 2.0 times 10^8 C B) q_2 = 5.0 times 10^8 C and q_3 = 1.0 times 10^8 C C) q_2 = 3.0 times 10^-8 C and q_3 = 3.0 times 10^8 C D) q_2 = 3.0 times 10^8 C and q_3 = 1.0 times 10^8 C E) q_2 = 1.0 times 10^8 C and q_3 = 3.0 times 10^8 C

Explanation / Answer

given that the charge on each plate of the capacitor c1 = 4.0*10-8c = 40*10-9c =40nc

let the upper plate of the capacitor c1 is positively charged and lower plate is negatively charged

as capacitor maintain equal and opposite charge on their plates, so +40nc charge on upper plate and -40nc charge on lower plate.

after being connected to two other capacitor c2 and c3, charge on capacitor c1 is 1.0*10-8c = 10nc

so, there is a decrease of (40-10) =30nc charge. hence this must be supplied to other two capacitors.

as c2 and c3 are in series, so they both recieve same charge. and according to our assumption, there will be +30nc chargeon their upper plate and -30nc charge on their lower plate.

hence the charge on positive plate of c2 and c3 are +30nc = 3.0*10-8c

so, option (c) is correct.

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