Question

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

• A. \( V = \dfrac{5q}{4\pi\varepsilon_0 r} \) and \( \vec{E} = \dfrac{5\sqrt{3}q}{8\pi\varepsilon_0 r^2}\,\hat{r} \)
• B. \( V = 0 \) and \( \vec{E} = 0 \)
• C. \( V = \dfrac{5q}{4\pi\varepsilon_0 r} \) and \( \vec{E} = \dfrac{5q}{4\pi\varepsilon_0 r^2}\,\hat{r} \)
• D. \( V = \dfrac{5q}{4\pi\varepsilon_0 r} \) and \( \vec{E} = 0 \)

Hint:

At the center of a symmetric charge distribution, electric fields cancel vectorially, but electric potential always adds.

Answer 1

The Correct Option is D

Step 1: Electric potential at the center.
Electric potential is a scalar quantity and hence adds algebraically. Each charge \(q\) is at the same distance \(r\) from the center \(O\). Therefore, potential due to one charge is \[ V_1 = \frac{q}{4\pi\varepsilon_0 r}. \] For five identical charges, \[ V = 5V_1 = \frac{5q}{4\pi\varepsilon_0 r}. \]
Step 2: Electric field at the center.
Electric field is a vector quantity. The electric fields due to the five charges at the center are equal in magnitude and symmetrically distributed in direction. Because of the regular pentagonal symmetry, the vector sum of the electric fields cancels out: \[ \vec{E} = 0. \]
Final Answer: \[ \boxed{V = \dfrac{5q}{4\pi\varepsilon_0 r} \quad \text{and} \quad \vec{E} = 0} \]