Question

Sixteen point charges each of charge \( q \) are placed on the circumference of a circle of radius \( r \) with equal angular spacing. If one of the charges is removed, then the net electric field at the centre of the circle is \( \varepsilon_0 \) (where \( \varepsilon_0 \) is the permittivity of free space). What is the net electric field at the centre?

• A. \( \frac{14q}{4\pi \varepsilon_0 r^2} \)
• B. \( \frac{16q}{4\pi \varepsilon_0 r^2} \)
• C. \( \frac{q}{4\pi \varepsilon_0 r^2} \)
• D. \( \frac{q}{2\pi \varepsilon_0 r^2} \)

Hint:

In symmetrical charge configurations like this one, removing a single charge disrupts the symmetry, leading to a net electric field. In such cases, always consider the effect of the remaining charges and the distance from the center.

Answer 1

The Correct Option is C

In the given scenario, 16 point charges are symmetrically placed on the circumference of the circle. Due to the symmetry, the electric field due to each charge at the center will cancel out each other when all the charges are present. The net electric field at the center would be zero.
When one charge is removed, there will be an imbalance in the charges. The remaining charges will no longer cancel each other perfectly, and there will be a net electric field due to the remaining 15 charges.
The net electric field due to a point charge at the center of the circle is given by: \[ E = \frac{q}{4\pi \varepsilon_0 r^2} \] Since there are 15 charges contributing to the net field, we multiply the field of one charge by the number of charges (15 remaining): \[ E_{\text{net}} = \frac{q}{4\pi \varepsilon_0 r^2} \] Thus, the net electric field is \( \frac{q}{4\pi \varepsilon_0 r^2} \).