Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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# A Charge of + 2.0 × 10 − 8 C is Placed on the Positive Plate and a Charge of − 1.0 × 10 − 8 C on the Negative Plate of a Parallel-plate Capacitor of Capacitance 1.2 × 10 − 3 Uf . - Physics

ConceptElectric Potential Difference

#### Question

A charge of +2.0 xx 10^-8 C  is placed on the positive plate and a charge of -1.0 xx 10^-8 C on the negative plate of a parallel-plate capacitor of capacitance 1.2 xx 10^-3  "uF" . Calculate the potential difference developed between the plates.

#### Solution

The charge on the positive plate is q1 and that on the negative plate is q2.
Given :

q_1 = 2.0 xx 10^-8 C

q_2 = -1.0 xx 10^-8 C

Now ,

Net charge on the capacitor = ((q_1 - q_2))/2 = 1.5 xx 10^-8 C

The  potential difference developed between the plates is given by q = VC

⇒ V = (1.5 xx 10^-8)/(1.2 xx 10^-9) = 12.5 V

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Solution A Charge of + 2.0 × 10 − 8 C is Placed on the Positive Plate and a Charge of − 1.0 × 10 − 8 C on the Negative Plate of a Parallel-plate Capacitor of Capacitance 1.2 × 10 − 3 Uf . Concept: Electric Potential Difference.
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