A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor .
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Hint:- Write the expression of potential difference between the plates when the slab is inserted and hence find capacitance by using the formula,c=vq
Step 1: Write the expression of capacitance before the slab is inserted.
Initially, when there is a vacuum between the two plates, the capacitance of the capacitor is
where A is the area of parallel plates
Suppose that the capacitor is connected to a battery, an electric field E0 is produced
Step 2: Write the expression of potential difference after the slab is inserted between the plates of the capacitor
if we insert the dielectric slab of thickness t=d/2 the electric field is reduced to E
Now, the gap between plates is divided in two parts, for distance t=2dthere is electric field E and for the remaining distance (d-t) the electric field is E0
If V be the potential difference between the plates of the capacitor, then
Step 3:Calculate capacitance after dielectric slab is inserted.
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