Electrostatics Question 277
Question: In an isolated parallel plate capacitor of capacitance C, the four surface have charges $ Q _{1} $ , $ Q _{2} $ , $ Q _{3} $ and $ Q _{4} $ as shown. The potential difference between the plates is [IIT-JEE 1999]
Options:
A) $ 400V $
B) $ 800\ V $
C) $ 1200\ V $
D) $ 1600\ V $
Show Answer
Answer:
Correct Answer: B
Solution:
Charge on capacitor A is given by
$ Q _{1}=15\times {{10}^{-6}}\times 100=15\times {{10}^{-4}}C $
Charge on capacitor B is given by $ Q _{2}=1\times {{10}^{-6}}\times 100={{10}^{-4}}C $
Capacity of capacitor A after removing dielectric $ =\frac{15\times {{10}^{-6}}}{15}=1\mu F $
Now when both capacitors are connected in parallel their equivalent capacitance will be Ceq $ =1+1=2\mu F $
So common potential $ =\frac{(15\times {{10}^{-4}})+(1\times {{10}^{-4}})}{2\times {{10}^{-6}}}=800V. $
