Answer to Question #138547 in Electric Circuits for Vakel

Question

Three charged particles are located at the corners of an equilateral triangle with sides 2.00C, -4.00C, 7.00C subtended by an angle of 60.0 degree at one end . Calculate the total electric force on the 7.00C charge.

Answer 1

Let us assume that side of the equilateral triangle is 1m.

Let $q_1 = 7.00C, q_2=2.00C, q_3 = -4.00C$

Force on 7C due to 2C is given by, $F_1 = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2} = \frac{9\times 10^9 \times 7 \times 2}{1^2} = 126\times 10^9 N$

Force on 7C due to -4C is given by, $F_2 = \frac{1}{4\pi \epsilon_0} \frac{q_1q_3}{r^2} = \frac{9\times 10^9 \times 7 \times 4}{1^2} = 252\times 10^9 N$

Since, these forces are acting at angle of $120^{\circ}$

Then, resultant of the forces is given by,

$F = \sqrt{F_1^2+F_2^2+2F_1F_2cos\theta} = \sqrt{(126\times 10^9)^2+(252\times 10^9)^2+2(252\times 10^9)(126\times 10^9)cos120}$

$F = (126\times 10^9)\sqrt{1+4+4cos120} = (126\times 10^9)\sqrt{1+4-2} = (126\sqrt{3}\times 10^9) N$

$F = 218.24 \times 10^9 N$