Question

A triangle with equal sides of length 14 cm has -2.5-nC charged objects at each corner. Determine the direction of the electrical force exerted on the object at the top corner due to the two objects at the horizontal base of the triangle.
(A) vertically down
(B) Vertically upward
(C) horizontally rightward
(D) horizontally leftward

Answer 1

Answer:

(B) Vertically upward

Explanation:

r = Side of triangle = 14 cm

q = Charge = -2.5 nC

Electrical force is given by

$F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^9\times (2.5\times 10^{-9})^2}{0.14^2}\\\Rightarrow F=2.86671\times 10^{-6}\ N$

For the top charge

Net force on both charges is given by

$F_n=2Fcos\theta\\\Rightarrow F_n=2\times 2.86671\times 10^{-6}\times cos30\\\Rightarrow F_n=4.96529\times 10^{-6}\ N$

The net force acting on the top charge is $4.96529\times 10^{-6}\ N$

Here the forces are symmetrical hence, the net force is along +y axis i.e., upward