Discussion Forum

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 = $\frac{1}{4 \pi \epsilon_{0}} \frac{q}{L^{2}}$, which of the following statements(s) is(are) correct.

Answer:

Explanation:

Resultant of 2K and 2K (at $120^{0}C)$ is also 2K towards 4K. Therefore , net electric field is 6K

 (b)   $V_{0}=\frac{1}{4 \pi \epsilon_{0}}$

      =$\frac{1}{4 \pi \epsilon_{0} L} (q_{A}+.....+q_{E})$

=0

 Because     $q_{A}+q_{B}+q_{C}+q_{D}+q_{E}+q_{F}=0$

 (c) only line PR , potential  is same (=0)