CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
KV
loader
C
VK
loader
D
3KV
loader

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image