Question

# Six charges are placed at the vertices of a regular hexagon as shown in the figure. Find the electric field on the line passing through O and perpendicular to the plane of the figure as a function of distance x from point O:

A
0
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B
4
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C
1
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D
2
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Solution

## The correct option is B 0The electric field due to a charge is given by →E=14πϵ0Qr3→rThus, the electric field at P which is located at a distance x from O is given by superposition of fields due to individual charges.Let k=14πϵ0 and d=√a2+x2Then →EA=kQd3(x^k−→OA)Similarly, →EB=−kQd3(x^k−→OB) , →EC=kQd3(x^k−→OC),→ED=−kQd3(x^k−→OD),→EE=kQd3(x^k−→OE),→EF=−kQd3(x^k−→OF)Thus, →E=kQd3(x^k(1−1+1−1+1−1)+(→OA+→OC+→OE)−(→OB+→OD+→OF))=0

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