The capacitance of a capacitor becomes 76 times its original value if a dielectric slab of thickness, t=23d is introduced in between the plates. d is the separation between the plates. The dielectric constant of the dielectric slab is :
1411
1114
711
117
A
711
B
1411
C
1114
D
117
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Solution
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Now Ceff is series combination of l0Ax,l0Ad−t−xandkl0At
1Ceff=xl0A+d−t−xl0A+tkl0A
⇒Ceff=kl0A(d−t)k+t
⇒ given Ceff=76⋅l0Ad
76⋅l0Ad=kl0A(d−23d)k+23d
⇒76=3k(13)k+23
⇒7k+14=18k
⇒k=1411
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