An air filled parallel plate condenser has a capacity of 2 pF. The separation of the plates is doubled and the interspace between the plates is filled with wax. If the capacity is increased to 6 pF, the dielectric constant of wax is :

Question

An air filled parallel plate condenser has a capacity of 2 pF. The separation of the plates is doubled and the interspace between the plates is filled with wax. If the capacity is increased to 6 pF, the dielectric constant of wax is : (A) 2 (B) 3 (C) 4 (D) 6

Tardigrade Admin · Accepted Answer

C text old =(ε 0 A / d )=2 pF arrow (i) C text new =( K ε 0 A /2 d )=6 pF arrow (ii) By Equating (i) and (ii) ( K /2)(2) pF =6 pF ⇒ K =6

Q. An air filled parallel plate condenser has a capacity of $2 pF$ . The separation of the plates is doubled and the interspace between the plates is filled with wax. If the capacity is increased to $6 pF$ , the dielectric constant of wax is :

2

3

4

6

$C_{old} = \frac{ε _{0} A}{d} = 2 pF \to$ (i)
$C_{new} = \frac{K ε _{0} A}{2 d} = 6 pF \to$ (ii)
By Equating (i) and (ii)
$\frac{K}{2} (2) pF = 6 pF$
$\Rightarrow K = 6$

Q. An air filled parallel plate condenser has a capacity of 2pF. The separation of the plates is doubled and the interspace between the plates is filled with wax. If the capacity is increased to 6pF, the dielectric constant of wax is :

2

3

4

6

Cold ​=dε0​A​=2pF→ (i) Cnew ​=2dKε0​A​=6pF→ (ii) By Equating (i) and (ii) 2K​(2)pF=6pF ⇒K=6

Q. An air filled parallel plate condenser has a capacity of $2 pF$ . The separation of the plates is doubled and the interspace between the plates is filled with wax. If the capacity is increased to $6 pF$ , the dielectric constant of wax is :

$C_{old} = \frac{ε _{0} A}{d} = 2 pF \to$ (i)
$C_{new} = \frac{K ε _{0} A}{2 d} = 6 pF \to$ (ii)
By Equating (i) and (ii)
$\frac{K}{2} (2) pF = 6 pF$
$\Rightarrow K = 6$