Point charges of 4.75 µC and −1.50 µC are placed 0.350 m apart. (Assume the negative charge is located to the right of the positive charge. Include the sign of the value in your answers.)

(a) Where can a third charge be placed so that the net force on it is zero?
m to the right of the −1.50 µC charge
(b) What if both charges are positive?

m to the right of the 1.50 µC charge

Explanation:

Let the position of balance point be x distance from 1.5 µC on the right side . Balance point can not be on the left side or between the charges because in that case electric field by both the charges will be in the same direction .

For equilibrium of field

k x 4.75 µC / ( .35 +x )² -  k x  1.5  µC / x ² = 0

4.75 / ( .35 +x )² = 1.5 / x²

( .35 +x )² / x² = 4.75 / 1.5

( .35 +x )² / x² = 3.167

( .35 +x ) / x =  1.78

.35 + x = 1.78 x

.78 x = .35

x = .45 m ( + ve ) to the right of 1.50  µC

b ) When both the charges are positive , balance point will lie in between them because , electric field will be in opposite direction .

For equilibrium of field

k x 4.75 µC / ( .35 - x )² =   k x  1.5  µC / x ²

4.75 / ( .35 -x )² = 1.5 / x²

( .35 -x )² / x² = 4.75 / 1.5

( .35 -x )² / x² = 3.167

( .35 -x ) / x =  1.78

.35 -x = 1.78 x

2.78 x = .35

x = .126 m  ( - ve ) to the right of 1.50  µC

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6