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Question

(C) The electric potential $V$ at the centre is the algebraic sum of potentials due to individual charges: $V = \sum \frac{kq_i}{R} = \frac{k}{R} \sum

Vedclass · Accepted Answer

$V = 0, E = 0$

Vedclass · Answer

(C) The electric potential $V$ at the centre is the algebraic sum of potentials due to individual charges: $V = \sum \frac{kq_i}{R} = \frac{k}{R} \sum q_i$. Since there are five charges of $(+q)$ and five charges of $(-q)$, the net charge $\sum q_i = 5(+q) + 5(-q) = 0$. Thus, $V = 0$.For the electric field $E$ at the centre, each charge $(+q)$ produces a field $\vec{E}_0$ directed away from it, and each charge $(-q)$ produces a field $\vec{E}_0$ directed towards it. Due to the symmetry of the ten charges arranged at equal angular intervals of $36^\circ$, every charge has an diametrically opposite charge of equal magnitude but opposite sign. For example, charge $1$ $(+q)$ and charge $6$ $(-q)$ are diametrically opposite. The field due to charge $1$ is $\vec{E}_1$ (away from $1$) and the field due to charge $6$ is $\vec{E}_6$ (towards $6$). Since $1$ and $6$ are opposite, $\vec{E}_1$ and $\vec{E}_6$ point in the same direction and add up to $2\vec{E}_0$. This results in five such vectors of magnitude $2E_0$ separated by $72^\circ$ each. The vector sum of these five equal vectors separated by equal angles is zero. Therefore, $E = 0$.

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