Question

Condenser A has a capacity of 15μF when it is filled with a medium of dielectric constant 15. Another condenser B has a capacity 1μF with air between the plates. Both are charged separately by a battery of 100V. After charging, both are connected in parallel without the battery and the dielectric material being removed. The common potential now is

• A. 400V
• B. 800V
• C. 1200V
• D. 1600V

Answer 1

Answer: (B)

800V

Answer 2

Charge on capacitor A is given by $q_{1} = C_{1}\text{ x V}$

= (15 x 10^-6) (100) = 15 x 10^-4C

Charge on capacitor B is given by q₂ = C₂ x V

= (1 x 10^-6) (100) = 10^-4C

Capacity of condensers A after removing dielectric

$C' = \left( \frac{C_{1}}{K} \right) = \left( 15\text{ x }\frac{\text{1}\text{0}^{\text{-6}}}{15} \right) = 1$μF

Now when both condenser are connected in parallel their capacity will be

1μF + 1μF = 2μF

Common potential V = $\frac{q}{C}$

=$\frac{\left( 15\text{ x 1}\text{0}^{\text{-4}} \right) + (1\text{ x 1}\text{0}^{\text{-4}})}{\text{2 x 1}\text{0}^{\text{-6}}} = 800V$