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# Solution to FMDC Paper

FMDC 2012 Paper

Q10. If in a parallel plate capacitor we insert a metal sheet of half the thickness as compared with the spacing between the plates of the capacitor, the capacitance becomes...?
A) C/4
B) C/2
C) 2 C
D) 4 C
E) 0.5 C

Solution :
When the metal plate is inserted we get the following situation:

The red box shows a metal sheet of thickness d/2. The green colored plus and minus signs represent the charges on the plate of capacitors. When the metal plate is inserted the negative charges on the plate are attracted by the positive plate and positive charge are attracted by a negative plate, as shown above. This gives us two capacitors with the distance d/4 between them.
The original capacitance was:
C = ∈A/d
New capacitance of each capacitor will be :
C' = ∈A/d/4 = 4 ∈A/d = 4C
Now we have to identical capacitors in series, so there equivalent capacitance will be 2C .

1/Ceq = 1/C1 + 1/C2 for series combination of capacitors.