Question

Three equal positive charge of 10 statC are located at the vertices of an equilateral triangle of 2cm leg. What is the magnitude of the force acting on each charge.

Answer 1

Let us determine the force on 1st charge (at vertex A of the triangle) due to 2nd (vertex B) and 3rd (vertex C) charges. Due to the symmetry of triangle, the resultant force will be directed along the line connecting the center O of the triangle and the vertex A.

We know that the electric force for two equal charges in SI system is

$F = k\dfrac{q\cdot q}{r^2} = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{q^2}{r^2}$ . For both 2nd and 3rd charges this force should be projected onto the line OA, so the components of forces will be

$F\cos \dfrac{60^\circ}{2} = F\cos 30^\circ.$ Therefore, the total force on 1st charge due to 2nd and 3rd charges is

$2F\cos30^\circ = \dfrac{2}{4\pi\varepsilon_0}\cdot\dfrac{q^2}{r^2}\cos30^\circ = \dfrac{\sqrt{3}}{4\pi\varepsilon_0}\cdot\dfrac{q^2}{r^2}.$

Now let us calculate the force. If we measure charges in statcoulombs and distance in cm, then in Gaussian-cgs system the force will take the form

$F = \dfrac{q^2}{r^2}$ , so in this system the total force takes form $2F\cos30^\circ = {\sqrt{3}}\cdot\dfrac{q^2}{r^2} = {\sqrt{3}}\cdot\dfrac{10^2}{2^2} \approx 43.3\,\mathrm{dynes}.$