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(B) The electric field at the centre of a regular hexagon due to a charge $q$ at a vertex is given by $E = \frac{kq}{r^2}$,where $r$ is the distance f

Vedclass · Accepted Answer

$2$

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(B) The electric field at the centre of a regular hexagon due to a charge $q$ at a vertex is given by $E = \frac{kq}{r^2}$,where $r$ is the distance from the vertex to the centre.For the electric field at the centre to be zero,the vector sum of the electric fields produced by all individual charges must be zero.In case $(1)$,all charges are equal,so the fields from opposite vertices cancel each other out,resulting in $E_{net} = 0$.In case $(2)$,two charges are $-q$ instead of $q$. This creates an unbalanced field. Specifically,the fields from the $-q$ charges point towards the charges,while the fields from the $q$ charges point away. This results in a non-zero resultant electric field $(E_{net} 
eq 0)$.In case $(3)$,opposite vertices have equal charges ($2q, q, 2q$ on one side and $2q, q, 2q$ on the other),so the fields cancel out,resulting in $E_{net} = 0$.In case $(4)$,the arrangement is symmetric such that the vector sum of the electric fields at the centre is zero $(E_{net} = 0)$.Therefore,the electric field at the centre is not zero only in case $(2)$.

The figures below show regular hexagons with charges at the vertices. In which of the following cases is the electric field at the centre not zero?

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