Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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When 1⋅0 × 1012 Electrons Are Transferred from One Conductor to Another, a Potential Difference of 10 V Appears Between the Conductors. Calculate the Capacitance of the Two-conductor System. - Physics


When 1⋅0 × 1012 electrons are transferred from one conductor to another, a potential difference of 10 V appears between the conductors. Calculate the capacitance of the two-conductor system.

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As 1.0 × 1012 electrons are transferred from one conductor to another, the conductor to which the electrons are transferred becomes negatively charged and the other conductor becomes positively charged.

Now , 

Magnitude of the net charge on each conductor, `Q = (1.0 xx 10^12) xx (1.6 xx 10^-19) C = 1.6 xx 10^-7 C`

Magnitude of the potential difference between the conductors, V = 10 V

The capacitance C is the ratio of the magnitude of the charge on either conductor to the magnitude of the potential difference between the conductors.

`C = Q/V`

⇒ `C = (1.6 xx 10^-7)/10 = 1.6 xx 10^-8 F`

Hence, the value of the capacitance of the given two conductor systems is `1.6 xx 10^-8 F` .

Concept: Capacitors and Capacitance
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HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 9 Capacitors
Q 1 | Page 165
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