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Two identical metal plates are given pos...

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

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

The correct Answer is:
D
Electric field with in the plates `E=E_(Q1) +E_(Q2) , E=E_1-E_2=(Q_1)/(2Aepsilon_0)-(Q_2)/(2Aepsilon_0)`
`E=(Q_1-Q_2)/(2Aepsilon_0)` `therefore` Potential between the plates
`V_A-V_B=Ed=((Q_1-Q_2)/(2A epsilon_0))d=(Q_1-Q_2)/(2((Qepsilon_0)/d))=(Q_1-Q_2)/(2C)`
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Knowledge Check

  • A capacitor or capacitance C_(1) is charge to a potential V and then connected in parallel to an uncharged capacitor of capacitance C_(2) . The fianl potential difference across each capacitor will be

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