The capacitance of a parallel plate capacitor with plate area A and separation d, is C. The space between the plates is filled with two wedges of dielectric constant K 1 and K 2 respectively (figure). Find the capacitance of the resulting capacitor.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
C R = 
ln 
where C = 
Sol.
take ‘dx’ element at x distance 
= 
; y = 
dC 1 = 
= 
dC 2 = 
= 
dC 1 and dC 2 are in series, so their equivalent
dC = 
= 
Now, we can consider there parallel slabs to the parallel in circuit combination
C eq. = dC 1 + dC 2 + dC 3 + dC 4
= 
= 
= 
{ln λ K 1 – ln K 2 λ }
C = 
ln 
= 
λ n 
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