The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery of voltage, V. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is :-

Question

The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery of voltage, V. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is :- (A) (V/K + n) (B) V (C) ((n + 1) V/(K + n)) (D) (nV/K + n)

Tardigrade Admin · Accepted Answer

Total charge of the system = CV + nCV = (n + 1)CV After the insertion of dielectric of constant K New potential (common) VC = (total charge/total capacitance) = (( n + 1) CV/KC + nC) = ((n + 1) V/K + n)

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

V

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

$\frac{nV}{K + n}$

Total charge of the system = $C V + n C V$
$= (n + 1) C V$
After the insertion of dielectric of constant $K$
New potential (common)
$V_{C} = \frac{t o t a l c ha r g e}{t o t a l c a p a c i t an ce}$
$= \frac{( n + 1 ) C V}{K C + n C} = \frac{( n + 1 ) V}{K + n}$

K+nV​

V

(K+n)(n+1)V​

K+nnV​

Total charge of the system = CV+nCV =(n+1)CV After the insertion of dielectric of constant K New potential (common) VC​=totalcapacitancetotalcharge​ =KC+nC(n+1)CV​=K+n(n+1)V​

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

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

$\frac{nV}{K + n}$

Total charge of the system = $C V + n C V$
$= (n + 1) C V$
After the insertion of dielectric of constant $K$
New potential (common)
$V_{C} = \frac{t o t a l c ha r g e}{t o t a l c a p a c i t an ce}$
$= \frac{( n + 1 ) C V}{K C + n C} = \frac{( n + 1 ) V}{K + n}$