Question

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Answer 1

The initial capacity of capacitor is given 5 microfarad

After inserting dielectric the capacity of a capacitor become 10 micro Farad

So C1=5 microfarad

C2=10 microfarad

Q=C*V

Q1=5 microcoulomb

Q2=10 microcoulomb

Hence additional charge flow equal to

∆Q=Q2-Q1

=(10-5) = 5 microcoulomb

Answer 2

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Answer- 5 uC

Above ur Explaination ✌✌

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