Question

Six point charges are kept at the vertices of a regular hexagon of side $L$ and centre $O$ as shown in the figure. Given that $K=\frac{1}{4 \pi \varepsilon_0}\frac{q}{L^2},$ which of the following statements(s) is(are) correct.

• A. The electric field at O is 6K along OD
• B. The potential at O is zero
• C. The potential at all points on the line PR is same
• D. The potential at all points on the line ST is same

Answer 1

Answer: (A), (B), (C)

(a) Resultant of 2K and 2K (at $120^\circ$) is also 2K towards
4A .Therefore, net electric field is 6A.
(b) $V_0=\frac{1}{4 \pi\varepsilon_0} \bigg[\frac{q_A}{L}+\frac{q_B}{L}+\frac{q_C}{L}+\frac{q_D}{L}+\frac{q_E}{L}+\frac{q_F}{L}\bigg]$
$ =\frac{1}{4 \pi\varepsilon_0} (q_A+...+q_F) $
$ = 0$
Because $ {q_A}+{q_B}+{q_C}+{q_D}+{q_E}+{q_F}=0 $
(c) Only line PR. potential is same (= 0).