1 Three charges of equal magnitude q reside at the corners of an equilateral tri

Question

1 Three charges of equal magnitude q reside at the corners of an equilateral triangle ofside length a Two of the charges are negative, and the other is positive, as shown in Figure 23.40. (a) Find the magnitude and direction of the electric field at point P, midway between the negative charges, in terms of k, q, and a. (b) Where must a-4a charge be placed so that any charge located at point P will experience no net elec- trostatic force (F 0)? In (b) let the distance be- tween the q charge and point Pbe one meter. Figure 23.40 a/2 a/2

Explanation / Answer

The magnitude of electric field is given by (E) = (kq) / (r2) .......where, q is charge and r is distance from the charge to the point where field has to be measured.

The field is along the line joining this charge to the point and direction is away from the charge in case of positive charge and towards it in case of negative one.

A)

In our case since the point P is at the midpoint (a/2 distance) of both the negative charges, so the resultant field is zero due to its equal magnitude but opposite direction.

The only field experienced at point P is due to positive charge so the direction is verticaly downwards.

Ep = kq / [ a sin(600) ]2 = (4kq) / (3a2) or 1.3333 kq/a2

Magnitude of field at P is (4kq) / (3a2) or 1.3333 kq/a   and direction is verticaly downwards (away from the charge +q)

B)

Let charge at P be Q

Force (F) = QEp where Q is charge at point P and E is field due to other charges at P.

therefore, force on P due to this three charges is F = Qkq . since, distance between +q and P is 1 meters.

Its direction is in the direction of field which is downwards.

For making net Force at P (Fp) = 0, by placing -4q charge one has to make field at point P to be zero.

this can happen by placing the -4q charge above the the ponit P along the line joining P and positive charge. So, that its direction becomes upward.

After placing -4q charge at a distance x from P along the line joining P and positive charge.

Field at P (Ep) = kq - 4kq/x2 since, distance between +q and P is 1 meters.

0 = kq - 4kq/x2

0 = 1 - 4/x2

x2 = 4

x = ± 2 m

since we have already discused that the -4q charge should be place above P so the position of -4q charge is 2m verticaly above the point P and 1m verticaly above the +q charge.

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