Answer to Question #135096 in Organic Chemistry for Flame

Question

Two particles having charge of +0.540 nC and +8.64 nC are separated by a distance of 1.90 m.

(a) At what point along the line connecting the two charges is the net electric field due to the two charges equal to zero?

(b) Where would the net electric field be zero if one of the charges were negative?

Answer 1

(a) Electric field is directed from the positive charge, so on the line between two positive charges the summary field becomes zero at the point where the fields have equal magnitudes. The magnitude of electric field E is proportional to: E ~ q/r², where q is charge and к is radial distance from the charge.

If x is the point of zero electric field, x is also the distance from the charge q1 = +8.64 nC, when (1.90 - x) is the distance from the charge q2 = +0.540 nC. The magnitudes of electric fields will be equal at x and the total electric field will be zero, if:

8.64/x² = 0.540/(1.90 - x)²

Solving the square equation gives x = 1.52 m.

Answer: net electric field is zero at the point of 1.52 m from the charge +8.64 nC, or 1.90-1.52 = 0.38 m from the charge +0.540 nC.

(b) The net electric field between positive and negative charges would be zero only at an infinite distance aside from the electric dipole.