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

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

Step-by-step 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=(\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}'=\frac{K\varepsilon _oA}{{d}'}\Delta V`

`{Q}'=(4.2\varepsilon _oA)/(2d)\Delta V` ----2

Now from equation 1 and 2

`\frac{{Q}'}{Q}=(4.2)/(2)`

So

`{{Q}'}=2.1Q`