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

To solve the problem, we need to determine the electric potential \( V \) and the electric field \( \vec{E} \) at the center \( O \) of a regular pentagon formed by five positive charges, each of charge \( q \).

Electric Potential \( V \) 

The electric potential at a point due to a point charge is given by the formula:

\(V = \dfrac{q}{4\pi\varepsilon_0 r}\)

where:

• \(q\) is the charge,
• \(r\) is the distance from the charge to the point where the potential is being calculated,
• \(\varepsilon_0\) is the permittivity of free space.

Since there are five identical charges placed symmetrically at the vertices of the pentagon, the potential at the center is the sum of potentials due to each charge. Therefore, the total potential is:

\(V = 5 \times \dfrac{q}{4\pi\varepsilon_0 r} = \dfrac{5q}{4\pi\varepsilon_0 r}\)

Electric Field \( \vec{E} \)

The electric field due to a point charge is a vector quantity given by:

\(\vec{E} = \dfrac{q}{4\pi\varepsilon_0 r^2}\,\hat{r}\)

However, because the charges are symmetrically placed at the vertices of the pentagon, their individual electric fields at the center will cancel out due to symmetry. Each pair of opposing charges creates equal and opposite electric field vectors that cancel each other.

Hence, the net electric field at the center \( O \) is:

\(\vec{E} = 0\)

Conclusion

Based on the calculations and symmetry of the problem:

• The electric potential at the center is \(V = \dfrac{5q}{4\pi\varepsilon_0 r}\)
• The electric field at the center is \(\vec{E} = 0\)

Therefore, the correct answer is:

\( V = \dfrac{5q}{4\pi\varepsilon_0 r} \) and \( \vec{E} = 0 \)