Two particles A and B having charges ` 8.0 xx 10 ^(-6) C` and `-2.0 xx 10^(-6) C` respectively are held fixed with a. separation of 20cm. Where should a third charged. particle be placed so that it dose not experience a net electric force?.

Options

A. 20 cm
B. 40 cm
C. 60 cm
D. 45 cm

Answer

A

Explanation

As the net electric force on C should be equal to
zero, the force due to A and B must be opposite in
direction. Hence, the particle should be placed on the
line AB. As A and B have charges of opposite signs, C
cannot be between A and B. Also, A has larger
magnitude of charge then B. Hence, C should be placed closer to B then A. The situation is shown in figure

Suppose `BC = x and the charge on C is Q.`
The force due to `A = ((8.0xx10^(-6)C)Q)/(4piepsilon_0(20cm +x)^2)`
The force due to `B = ((2.0xx10^(-6)C)Q)/(4piepsilon_0x^2)`
They are oppositely directed and to have a zero
resultant, they should be equal in magnitude. Thus,
`(8)/((20cm+x)^2) = (2)/(x^2)`
or `(20cm+x)/(x) = 2,` giving x = 20cm.