Capacitance of a capacitor becomes 43 times its original value if a dielectric slab of thickness t=d2 is inserted between the plates (d = separation between the plates). Then the dielectric constant of the slab is
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Solution
From given Cf=43Ci
After inserting dielectric the capacitor can be redrawn as shown below
C1=ε0Ad/2;C2=Kε0Ad/2
Now, Ceq=C1×C2C1+C2
On solving we get Ceq=2KεoAd(K+1)=Cf
From given condition , 2Kε0Ad(K+1)=43ε0Ad
And on solving the above equation, we get K=2.