
Six point charges each of the magnitude \(Q\) are placed at the vertices of a regular hexagon of side \(a\)…

- \(k \times \frac{4 Q a}{x^3}\)
- \(k \times \frac{2 Q a}{x^3}\)
- \(k \times \frac{8 Q a}{x^3}\)
- 0
Solution

Let \(O K=x\) due to \(A B\), force at the point \(K\) is only along \(A B\) \(\text {i.e., } F_K=\frac{2 k Q a}{\left(a^2+x^2\right)^{\frac{3}{2}}}\) \(\therefore\) There are three such \(F_K^{\prime} s\) so the net resultant of these three, \(\begin{aligned} & F_{\text {net }}=\frac{4 k Q a}{\left(a^2+x^2\right)^{\frac{3}{2}}}=\frac{4 k Q a}{x^3\left(1+\frac{a^2}{x^2}\right)^{\frac{3}{2}}} \\ & F_{\text {net }}=\frac{4 k Q a}{x^3} \quad\left(\therefore \frac{a^2}{x^2}=0\right),(x > > a) \end{aligned}\)
Asked in: AP EAMCET 2019 (22 Apr Shift 1)