Question

A parallel plate capacitor with capacitance C is connected to a power supply AV, acquiring a charge on its plates. While it is still connected to the power supply, the plate separation is doubled and a dielectric of constant K = 4.20 is inserted into the gap. What is the new charge on the plates in the final state? a) Q=0.4769 b) Q=2.109 c) Q=4.20g d)

Answer 1

Answer:

b)${{Q}'}=2.1Q$

Explanation:

Given that

The power source still connects it means that the voltage difference will be same in above both condition

As we know that

$Q=C\Delta V$

Or we can say that

$Q=\dfrac{\varepsilon _oA}{d}\Delta V$    ----1

Where d is the distance between two plates ,A is the area.

When K=4.2 and distance become double

${Q}'=\dfrac{K\varepsilon _oA}{{d}'}\Delta V$

${Q}'=\dfrac{4.2\varepsilon _oA}{2d}\Delta V$    ----2

Now from equation 1 and 2

$\dfrac{{Q}'}{Q}=\dfrac{4.2}{2}$

So

${{Q}'}=2.1Q$