Answer to Question #138227 in Mechanics | Relativity for Basa

Question

Three point charges are arranged such that they take the shape of an isosceles triangle with sides 2.00C, -4.00C, 7.00C subtended by an angle of 60.0 degree at one end. Find(a) the magnitude and (b) the direction of the electric force on the particle at the origin.

Answer 1

Let the particles be arranged such that -4C is at origin.

Let 2C be placed on y-axis at a distance of 1m.

Let 7C be placed on x-axis at a distance of 1m.

Then force due to 2C charge on the charge at origin,

$F_1 = \frac{1}{4\pi \epsilon_0} \frac{2\times 4 }{1^2} = 72\times 10^9 N$ along y-axis.

Force due to charge 7C,

$F_2 = \frac{1}{4\pi \epsilon_0} \frac{4\times 7}{1^2} = 252\times 10^9 N$ along x-axis.

Since $F_1$ and $F_2$ are perpendicular to each other then resultant force will be,

$F = \sqrt{F_1^2 + F_2^2} = 262.084 N$

Direction is given by,

$\theta = tan^{-1}\frac{ 72 \times 10^9}{ 252 \times 10^9} = 15.9^{\circ}$ with the x-axis.