Question

The plates of a parallel plate capacitor are given charges +4Q and -2Q . The capacitor is then connected across an uncharged capacitor of same capacitance as first one (=C). Find the final potential difference bw plates of the first capacitor.

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Answer 1

Explanation:

Given that the capacitance of the two capacitors are,

C1=6 μF & C2=14 μF.

The first capacitor is charged by a potential 100 V. Therefore, the charge on the first capacitor is,

Q1=C1V1=6×10−6×100=6×10−4 C

Now, if this two capacitors kept in touch and separated, then let q1 & q2 are the charges on each capacitor. Therefore, according to the charge conservation,

q1+q2=6×10−4 C

When they kept in contact and separated, their potential becomes equal. So, the ratio of charges is equal to the ratio of capacitance of two capacitors.

q1q2=C1C2

⇒q1q2=614

⇒q1q2=37

Therefore,

q1+q2=6×10^−4

⇒q2(37+1)=6×10^−4

⇒q2=6×10−^4/(10/7)

⇒q2=6×7×10^−5

⇒q2=42×10^−5 C

So,

q1=37×42×10^−5

⇒q1=18×10^−5 C

Answer 2

Solution in attachment ⬆️

Hope it helps....