Join now for JEE/NEET and also prepare for Boards Join now for JEE/NEET and also prepare for Boards. Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
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! 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!
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
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
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -