Two parallel plate capacitors of capacitance C and 2C are connected in parallel and charged to a potential difference V0. The battery is then disconnected and the region between the plates of the capacitor C is completely filled with a material of dielectric constant 2 . The potential difference across the capacitors now becomes

Question

Two parallel plate capacitors of capacitance C and 2C are connected in parallel and charged to a potential difference V0. The battery is then disconnected and the region between the plates of the capacitor C is completely filled with a material of dielectric constant 2 . The potential difference across the capacitors now becomes (A) (V0/4) (B) (V0/2) (C) (3 V0/4) (D) V0

Tardigrade Admin · Accepted Answer

Initial state Solution

As both capacitors have the same capacitance total charge = C V0 + 2 C V0 = 3 C V0 It will be divided equaly so q1 = q2 = (3 C V0/2) & potential = (q/C) = ((. (3 C V0/2) .)/2 C) = (3 V0/4)

$\frac{V _{0}}{4}$

$\frac{V _{0}}{2}$

$\frac{3 V _{0}}{4}$

$V_{0}$

Initial state

As both capacitors have the same capacitance total charge $= C V_{0} + 2 C V_{0}$
$= 3 C V_{0}$
It will be divided equaly so $q_{1} = q_{2} = \frac{3 C V _{0}}{2}$ & potential $= \frac{q}{C} = \frac{( \frac{3 C V _{0}}{2} )}{2 C} = \frac{3 V _{0}}{4}$

4V0​​

2V0​​

43V0​​

V0​

Initial state As both capacitors have the same capacitance total charge =CV0​+2CV0​ =3CV0​ It will be divided equaly so q1​=q2​=23CV0​​ & potential =Cq​=2C(23CV0​​)​=43V0​​

$\frac{V _{0}}{4}$

$\frac{V _{0}}{2}$

$\frac{3 V _{0}}{4}$

$V_{0}$

Initial state

As both capacitors have the same capacitance total charge $= C V_{0} + 2 C V_{0}$
$= 3 C V_{0}$
It will be divided equaly so $q_{1} = q_{2} = \frac{3 C V _{0}}{2}$ & potential $= \frac{q}{C} = \frac{( \frac{3 C V _{0}}{2} )}{2 C} = \frac{3 V _{0}}{4}$