A spherical drop of liquid carrying charge, Q has potential `V_0` at its surface. If two drops of same charge and radius combine to form of single spherical drop, then the potential at the surface of new drop is (Assume, V = 0 at infinity.)

Options

A. `2 ^(1//3) V_(0)`
B. `4 ^(1//3) V_(0)`
C. `6 ^(1//3) V_(0)`
D. `2 ^(-1//3) V_(0)`

Answer

B

Explanation

Given,
For a spherical drop of liquid,
Charge = Q
Potential at its surface `= V_(0)`
Potential of drop, `V_(0) = (1)/( 4 pi epsi _(0)) (Q)/® " "…(1)`
let R be the radius of big single sherical drop obtained by combining two same charge drops.
Charge of new drop,` Q' =Q+ Q=2Q`
Radius of new drop = R.
Then,
Volume of new drop = Sum of volume of two individual drops
`4/3 pi R ^(3) = 4/3 pi r ^(3) + 4/3 pi r ^(3)`
`R^(3) = r ^(3) + r ^(3)`
`R ^(3) = 2r ^(3)`
`R = 2 ^(1//3) r`
`therefore` Potential at the surface of new drop, `V_(0)^(1) = (1)/( 4 pi epsi _(0)) (1)/(R)`
`V_(0) ^(1) = (1)/(4 pi epsi _(0)) (2Q)/(2 ^(1//3) r )`
`= (1)/(4 pi epsi_(0)) (Q)/(r) 2 ^(1- (1)/(3))`
`= V _(0) 2 ^((1)/(2)) (because` From equation (1)
`V_(0) ^(1) =V_(0) 4 ^(1//3)`
`therefore` Potential at teh surface of new drop, `V _(0) ^(1)= (1)/( 4 pi epsi _(0)) 1/R`
`V_(0) ^(1) = (1)/(4 pi epsi _(0)) (2Q)/(2 ^(1//3) r)`
`= (1)/(4 pi epsi _(0)) Q/r 2 ^(1- (1)/(3))`
`=V_(0) 2 ^((1)/(2)) (because` From equation (1)
`V _(0) ^(1) = V_(0) 4 ^(1//3)`