2. Fig. shows a charge =1.0 C is 10 cm from a charge = 2.0 C . At what point on the line joining the two charges is the electric field strength be zero ?
Add the two fields together and set them equal to zero. You will have one equation:
E1 + E2 = 0
Substitute for each field (E = 1/(4*pi*e0) * Q/r^2). Note that both Q and r are both different for each field, and that we don't know r for either equation. Furthermore, since the electric constant is the same for both fields, we can pull that term out and simplify the equation further to:
1/r1^2 - 2/r2^2 = 0. We're subtracting the second quantity because the 2nd field opposes the first field. If you add them together instead of subtract, you'll get a complex answer!
Now we relate r1 and r2. r1 is the distance from the first charge to the point. Since the total distance is 10 cm, r2 = 10 cm - r1. Substitute for r2 to get
1/r1^2 - 2/(10-r1)^2 = 0
One equation, one unknown. Solve for r1, which should be 4.14 cm.
It's always good to do a reality check. Since a distance of 4.14 cm puts the neutral field point closer to the first charge than the second, it's safe to say that the answer makes sense - you need to be farther away from the second charge to have a field as weak as the first charge.