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JEE Main 202622 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Evening Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2026 · 22 January · Evening Shift · Physics · Electrostatics

Five positive charges each having charge q are placed at the...

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:
Five positive charges each having charge q are placed at the vertices of a pentagon as shown in the figure. The

Options

  1. AV=5q4πεor and E⃗=53q8πεor2r^V=\frac{5 q}{4 \pi \varepsilon_{o} r} \text{ and } \vec{E}=\frac{5 \sqrt{3} q}{8 \pi \varepsilon_{o} r^{2}} \hat{r}
  2. BV=5q4πεor and E⃗=0V=\frac{5 q}{4 \pi \varepsilon_{o} r} \text{ and } \vec{E}=0
  3. CV=0 and E⃗=0V=0 \text{ and } \vec{E}=0
  4. DV=5q4πεor and E⃗=5q4πεor2r^V=\frac{5 q}{4 \pi \varepsilon_{o} r} \text{ and } \vec{E}=\frac{5 q}{4 \pi \varepsilon_{\mathrm{o}} r^{2}} \hat{r}

Correct answer

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

Step-by-step solution

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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