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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

Text Solution

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Let A be the area of each plate. When the two plates are placed d distance apart. Capacitance of parallel plate capacitor so formed is
`C = (in_(0) A)/(d)`
If `E_(1) and E_(2)` are electrons fields due to the two plates, then the net field between the plates.
`E = E_(1) - E_(2)`
`(sigma_(1))/(2 in_(0)) - (sigma_(2))/(2 in_(0)) = (1)/(2A in_(0)) (q_(1) - q_(2))`
P.d between the plates
`V = Ed = (1)/(2 in_(0) A) (q_(1) - q_(2)) xx d`
`V = (1)/((2 in_(0)A)/(d)) (q_(1) - q_(2)) = (q_(1) - q_(2))/(2C)`
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