Electrostatics

JEE Main level, single correct, medium

156 of 311

Question 156 of 311

Five positive charges each having charge qq are placed at the vertices of a pentagon as shown in the figure. The electric potential (V)(V) and the electric field (E⃗)(\vec{E}) at the center OO of the pentagon due to these five positive charges are:

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Answer: (B) V=5q4πεor and E⃗=0V=\frac{5 q}{4 \pi \varepsilon_{o} r} \text { and } \vec{E}=0

The electric potential at center O is the scalar sum of contributions from all 5 charges.
Each charge is equidistant from center, so V=5cdotkqr=5q4πϵ0rV = 5 cdot \frac{kq}{r} = \frac{5q}{4\pi\epsilon_0 r}.
The electric field is a vector sum of field vectors from all 5 charges.
Due to 5-fold rotational symmetry of a regular pentagon, the 5 field vectors point radially outward from center, equally spaced at 72° angles.
The vector sum of 5 equally spaced unit vectors is zero, so E⃗=0\vec{E} = 0.

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