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

Options

A. `(Q_(1)+Q_(2))//2C`
B. `(Q_(1)+Q_(2))//C`
C. `(Q_(1)-Q_(2))//C`
D. `(Q_(1)-Q_(2))//2C`

Answer

D

Explanation

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)`