Two identical parallel plate air capacitors are connected in series to a battery of emf . If one of the capacitor is completely filled with dielectric material of constant, then potential difference of the other capacitor will become

Question

Two identical parallel plate air capacitors are connected in series to a battery of emf . If one of the capacitor is completely filled with dielectric material of constant, then potential difference of the other capacitor will become (A) (K V/(K+1)) (B) (V/K+1) (C) ((K-1) V/K) (D) (V/K(K+1))

Tardigrade Admin · Accepted Answer

Both the capacitance will have same charge

V_{1} = \frac{q}{C}

and

V_{2} = \frac{q}{K C}

But

V_{1} + V_{2} = V

\Rightarrow \frac{q}{C} + \frac{q}{K C} = V \Rightarrow \frac{q}{C} \cdot \frac{K + 1}{K} = V

Potential difference across capacitor without dielectric is

V = \frac{q}{C} = \frac{K}{K + 1} V

Q. Two identical parallel plate air capacitors are connected in series to a battery of emf . If one of the capacitor is completely filled with dielectric material of constant, then potential difference of the other capacitor will become

$\frac{K V}{( K + 1 )}$

$\frac{V}{K + 1}$

$\frac{( K - 1 ) V}{K}$

$\frac{V}{K ( K + 1 )}$

Q. Two identical parallel plate air capacitors are connected in series to a battery of emf . If one of the capacitor is completely filled with dielectric material of constant, then potential difference of the other capacitor will become

(K+1)KV​

K+1V​

K(K−1)V​

K(K+1)V​

Both the capacitance will have same charge V1​=Cq​ and V2​=KCq​ But V1​+V2​=V ⇒Cq​+KCq​=V⇒Cq​⋅KK+1​=V​ Potential difference across capacitor without dielectric is V=Cq​=K+1K​V

$\frac{K V}{( K + 1 )}$

$\frac{V}{K + 1}$

$\frac{( K - 1 ) V}{K}$

$\frac{V}{K ( K + 1 )}$