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Electrostatic Potential and Capacitance

Energy Stored in a Capacitor

An insulated conductor initially free from charge is charged b

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An insulated conductor initially free from charge is charged by repeated contacts with a plate which after each contact is replenished to $a$ charge Q. If $q$ is the charge on the conductor after first operation prove that the maximum charge which can be given to the conductor in this way is $\frac{Qq}{Q - q}$ .

Name: Video solution 1: An insulated conductor initially free from charge is charged by repeated contacts with a plate which after each contact is replenished to a charge Q. If q is the charge on the conductor after first operation prove that the maximum charge which can be given to the conductor in this way is Q−qQq.
Uploaded: 2023-12-30T19:40:47.000Z
Duration: 4 min
Description: Video solution 1: An insulated conductor initially free from charge is charged by repeated contacts with a plate which after each contact is replenished to a charge Q. If q is the charge on the conductor after first operation prove that the maximum charge which can be given to the conductor in this way is Q−qQq.

Views: 5,248 students

Updated on: Dec 30, 2023

Book:

Electricity and Magnetism (DC Pandey)

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An insulated conductor initially free from charge is charged by repeated contacts with a plate which after each contact is replenished to

a

charge Q. If

q

is the charge on the conductor after first operation prove that the maximum charge which can be given to the conductor in this way is

\frac{Qq}{Q - q}

5 mins ago

Discuss this question LIVE

5 mins ago

Text solutionVerified

Solution Let

C_{1}

be the capacity of plate and

C_{2}

that of the conductor. After first contact charge on conductor is

q

. Therefore, charge on plate will remain

Q - q

. As the charge redistributes in the ratio of capacities.

\frac{Q - q}{q} = \frac{C _{1}}{C _{2}}

Let

q_{m}

be the maximum charge which can be given to the conductor. Then, flow of charge from the plate to the conductor will stop when,

V_{conductor} = V_{plate}

\frac{q _{m}}{C _{2}} = \frac{Q}{C _{1}} \Rightarrow q_{m} = (\frac{C _{2}}{C _{1}}) Q

Substituting

\frac{C _{2}}{C _{1}}

from Eq. (i), we get

q_{m} = \frac{Qq}{Q - q}

Hence Proved.

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Question Text	An insulated conductor initially free from charge is charged by repeated contacts with a plate which after each contact is replenished to $a$ charge Q. If $q$ is the charge on the conductor after first operation prove that the maximum charge which can be given to the conductor in this way is $\frac{Qq}{Q - q}$ .
Updated On	Dec 30, 2023
Topic	Electrostatic Potential and Capacitance
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 2
Upvotes	251
Avg. Video Duration	6 min

Text solutionVerified

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