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

Sum

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.

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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`

  Is there an error in this question or solution?
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APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 9 Capacitors
Q 30 | Page 167
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