Physics: Aparallel 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)

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Aparallel plate capacitor with capacitance c is

gjvyverman

01.01.2022

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

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

{{Q}'}=2.1Q

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