Question

Three point charges lie in a straight line along the y-axis. A charge of q1 = -10.00 µC is at y = 6.40 m, and a charge of q2 = -7.70 µC is at y = -3.60 m. The net electric force on the third point charge is zero. Where is this charge located?

Answer 1

Answer:

y coordinate where field is zero is  1.1 m

Explanation:

given data

charge q1 = -10.00 µC

y = 6.40 m,

charge q2 = -7.70 µC

y = -3.60 m

solution

we know that here Electric field E that is express as

Electric field E = $\frac{kQ}{r^2}$   ..............................1

and

Distance between charges will be

Distance between charges = 6.40 -  (-3.60) = 10 m

so here neutral point P is at  distance d from q1

so it is  distance (10 - d) from q2.  

so here

Field from q1 at P = $\frac{k(-10 \times 10^{-6})}{d^2}$

and

Field from q2 at P = $\frac{k(-7.70 \times 10^{-6})}{(10-d)^2}$

so here field is  in opposite directions and the resultant field = 0

so

$\frac{k(-10 \times 10^{-6})}{d^2} = \frac{k(-7.70 \times 10^{-6})}{(10-d)^2}$  

$\frac{10}{d^2} = \frac{7.70}{(10-d)^2}$

d =  5.326242 m

so y coordinate where field is zero is 6.40 - 5.32 = 1.1 m