1.

Two plates ( area =s ) charged to + q_(1)and +q_(2)(q_(2) lt q_(1))are brought closer to form a capacitor of capacitance C. The potential difference across the plates is

Answer»

`(q_(1)-q_(2))/(2C)`
`(q_(1)-q_(2))/( C)`
`(q_(1)-q_(2))/(4 C )`
`(2(q_(1)-q_(2)))/ ( C)`

SOLUTION :Answer (1)
Inner plates accumulate equal and opposite charge `Q=(q_(1)-q_(2))/(2)`and that actually MAKES the capacitor Now `V=(Q)/(C )`.


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