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 $K = \frac{1}{4 π ε_{0}} \frac{q}{L^{2}},$ which of the following statement(s) is (are) correct [2012]

Question

Smart Achievers · Accepted Answer

(1, 2, 3)

Here $O$

$∴$ $E_{O} = E_{A} + E_{D} + (E_{F} + E_{C}) \cos 60 ° + (E_{B} + E_{C}) \cos 60 °$

$= 2 K + 2 K + (K + K) (\frac{1}{2}) + (K + K) (\frac{1}{2})$

$= 2 K + 2 K + K + K = 6 K$

Electric potential at center, O

$V_{O} = \frac{1}{4 π ε_{0} L} [2 q + q + q - q - q - 2 q] = 0$

Potential at all points on the line PR is the same, but not on line ST.

PR is the perpendicular bisector (the equatorial line) for the electric dipoles AB, FE, and BC. Therefore, the electric potential will be zero at any point on PR.

Solution

Q.
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 $K = \frac{1}{4 π ε_{0}} \frac{q}{L^{2}},$ which of the following statement(s) is (are) correct [2012]

1 The electric field at O is 6K along OD

2 The potential at O is zero

3 The potential at all points on the line PR is same

4 The potential at all points on the line ST is same

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