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A charge of +2.0 x 10^-8 C is placed on the positive plate and a charge of -1.0 x 10^-8 C on the negative plate of a parallel plate capacitor of capacitance 1.2 x 10^-3 uF. Calculate the potential difference developed between the plates. Please add explanation!
Ans- 12.5 volt
Explanation- charge appearing on right side of plate-2 = charge appearing of left side of plate-1= (2 x 10-8-1 x10-8)/2
= 10-8/2 Coulomb
now the charge appearing on right side of plate-1 will be- 2 x 10-8-10-8/2 = 1.5x10-8coulomb
charge appearing on right side of plate-1 = charge of +ve plate of capacitor ( charge of left side of plate is not
taken as plate of charge be is does not give its electric field as well as potential inside the capacitor)
using Q = charge of +ve plate = CV
1.5x10-8coulomb = 1.2x 10-3 micro F x V= 1.2x 10-9 Farad x Vg
solving this- We get V=12.5 volt
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