Electrostatics Ques 42
- 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.
(2012)
(a) The electric field at $O$ is $6 K$ along $O D$
(b) The potential at $O$ is zero
(c) The potential at all points on the line $P R$ is same
(d) The potential at all points on the line $S T$ is same
Show Answer
Solution:
- (a,b,c)
Resultant of $2 K$ and $2 K$ (at $120^{\circ}$ ) is also $2 K$ towards $4 K$.Therefore, net electric field is $6 K$.
$$ \begin{aligned} V_{0} & =\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{A}}{L}+\frac{q_{B}}{L}+\frac{q_{C}}{L}+\frac{q_{D}}{L}+\frac{q_{E}}{L}+\frac{q_{F}}{L} \\ & =\frac{1}{4 \pi \varepsilon_{0} L}\left(q_{A}+\ldots+q_{F}\right)=0 \end{aligned} $$
Because $q_{A}+q_{B}+q_{C}+q_{D}+q_{E}+q_{F}=0$
(c) Only line PR, potential is same $(=0)$.
BETA
