Question

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

A
3VK+2
B
KV
C
VK
D
3KV

Solution

## The correct option is A $$\dfrac {3V}{K+2}$$Initial total charge of the system is: $$Q_i=Q_1+Q_2=CV+2CV=3CV$$When dielectric is inserted in $$C$$ so the capacitance becomes $$KC$$Final charge, $$Q_f=Q'_1+Q'_2=KC V'+2CV'=(K+2)CV'$$ where $$V'=$$ common potential after disconnected the battery.As battery is disconnected so total charge will remain unchanged. Thus, $$Q_i=Q_f \Rightarrow 3CV=(K+2)CV'$$$$\therefore V'=\dfrac{3V}{K+2}$$ Physics

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