Question

2.0 pC charges sit at the vertices of an equilateral triangle of a side .24 m.

a) Sketch the situation and show the electric field vectors and show the net electric field at the center of the triangle.

b) Find the electric field at the center of the triangle.

c) Repeat the calculation if one of the charges is replaced by a -2.0 pC charge.

Thank you!

Answer 1

part A:

partB :

at the centroid of the triangle, the field due to each charge is same and are inclined by 120 deg to each other.

Hence net EF due to same charges, at centroid is ZERO

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part C :

distance from each charge to centerof trainge is L/sqrt3

here l = 0.24 m  

so distance d = 0.24/sqrt3 = 0.138 m

E due to charge is E1 = Kq/d^2

E1 = (9e9 * 2e-12)/(0.138^2) = 0.945 N/C

E2 = (9e9 * 2e-12)/(0.138^2) = 0.945 N/C

these two are at 60 degs

so

E12^2 = E1^2 + E2^2 + 2 E1 E2 Cos 60

E12^2 = 0.945^2 + 0.945^2 + 2 * 0.945 * 0.945 * cos 60

E12 = 1.636 N/C this acts straight up

Edue to -ve charge is E- = kq/r^2

E- = (9e9 * 2e-12)/(0.138^2) = 0.945 N/C this acts opposite the effect of both charges

so ENet = 1.636 - 0.945 = 0.691 N/C