Question

In Fig. 2.102, three negative charges each amount -q and three positive charges +9, +qand +Q are placed alternately on the vertices of a regular hexagon. If the electric field intensity at o due to the charges at A, B, C, D, E is double that at O due to the charge Q, what will be the value of Q?

​

Answer 1

.

$electric \: field \: due \: to \: two \: charges \\ + q \: and \: - q \: on \: the \: axis \: of \: each \: line. \\ \\ E_O= \frac{2kq}{ {a}^{2} } + \frac{2kq}{ {a}^{2} } = \frac{4kq}{ {a}^{2} } \\ \blue{ \underline{Taking \: resultant \: of \: AD \: and \: BE} : } \\ E_R = \sqrt{ {E}^{2}_O + {E}^{2}_O + 2EE \cos 120°} \\ = \sqrt{2{E}^{2}_O +2 {E}^{2}_O \times ( - \frac{1}{2} )} \\ = E_O = \frac{4kq}{ {a}^{2} } \\ \blue{ \underline{Electric \: field \: on \: centre \: O \: on \: line \: CF}} \\ E'_O = \frac{2kq}{ {a}^{2} } + \frac{kQ}{ {a}^{2} } = \frac{k}{ {a}^{2} }(2q + Q) \\ E_{net} = 0 \\ E'_O + ( - E_{R}) = 0 \\ E'_O = E_{R} \\ \frac{k}{ {a}^{2} }(2q + Q) = \frac{4kq}{ {a}^{2} } \\ 2q + Q = 4q \\ \boxed{Q = + 2q} \\ \\ \red{\mathfrak{ \large{\underline{{Hope \: It \: Helps \: You}}}}} \\ \blue{\mathfrak{ \large{\underline{{Mark \: Me \: Brainliest}}}}}$