Two charges are arranged at the bottom of a triangle. The charges are +13nC for

Question

Two charges are arranged at the bottom of a triangle. The charges are +13nC for the first charge (q1) on the left side and –13 nC (q2) on the right side. The triangle is an equilateral triangle with each side being 13.0 cm. Point a is +6 cm from q1, point b is –4 cm from q1 along an x axis, and point c is at the vertex of the triangle.

What is the electric field at point a, b, and c?

I now put a charge of +6nC at points a, b, and c. What is the force on the new charge (consider the +6nC charge at each point individually)

Explanation / Answer

E due to a charge = kq / r^2

at point a ,

Ea = due to q1 + due ti q2

= 9 x 10^9 x 13 x 10^-9 / 0.06^2 + 9x 10^9 x 13 x 10^-9 / 0.07^2

= 56377.55 N/C

at point b ,

Eb = - kq1 / 0.04^2 + kq2 / (0.13 + 0.04)^2

= 9x 10^9 x 13 x 10^-9 x [ - 1 / 0.04^2 + 1 /0.17^2 ] = 69076.56 N/C

at C,

Ec = 2Kq / a^2 cos60 = 2 x 9 x 10^9 x 13 x 10^-9 x cos60 / 0.13^2 = 6923.08 N/C

F = qE

Fa = 56377.55 x 6 x 10^-9 = 3.38 x 10^-4 N

Fb = q Eb = 4.14 x 10^-4 N

Fc = 4.15 x 10^-5 N

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