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Two identical metal plates are given charges q1 and q2(<q1) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C. What will be the potential difference between the plates?


now...what i find in every book is that...they are solving this question by taking the electric field intensity to be σ/2ε[ss]0[/ss]...

but these are "metal" plates..that means charged conductors....so the elecric field should be σ/ε[ss]0[/ss]....and then the answer should come out to be(q1-q2)/C and not (q1 - q2)/2C

σ/ε tab hoga jab charge dono plate is having eqal and opp. charge

arre yar....par i hav always read.....EF near a charged conducting surface is σ/ε and is normal to the surface....while the field due to a non conducting plane sheet of charge has EF σ/2ε

n ya........is bat ko consider kar k to.....EF in between th plates of a capacitor should be 2σ/ε........nw i am really getting confused....

someone help yarrr.......


3 din bad paper hai...........