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Two identical metal plates are given poi...

Two identical metal plates are given poistive charges `Q_1` and `Q_2` `(ltQ_1)` respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C, the potencial difference between them is

A

`(Q_1+Q_2)//2C`

B

`(Q_1+Q_2)//C`

C

`(Q_1-Q_2)//C`

D

`(Q_1-Q_2)//2C`

Text Solution

Verified by Experts

d. For the capacitor to get charged upto `0.75V`m the charge on the plates should be
`q=CV=10^(-5)xx0.75`
`=0.75xx10^(-5)C`
Using the equation of charging of capacitor
`q-CE(1-e^(t//RC))`, we get
`0.75xx10^(-5)=10^(-5)=10^(-5)xx1.5[1-e^((t)/(10^(5)xx10^(-5)))]`
or `(1)/(2)=[1-e^(-1)]` or `e^(-1)=(1)/(2)`
Taking log on both sides, we get
`-t="IN" 2`
or `t=0.693s`
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