Question

Six charges (four \( +q \), two \( -q \)) are present at a circle of radius \( r \) and centered at the origin as shown. The electric field at the origin is.

• A. \( \frac{\sqrt{3q}}{4\pi \epsilon_0 r^2} \hat{i} \)
• B. \( \frac{\sqrt{3q}}{4\pi \epsilon_0 r^2} (-\hat{i}) \)
• C. \( \frac{\sqrt{3q}}{2\pi \epsilon_0 r^2} (-\hat{i}) \)
• D. \( \frac{\sqrt{3q}}{\pi \epsilon_0 r^2} \hat{i} \)

Hint:

When dealing with multiple charges, remember to calculate the vector sum of the individual electric fields. Symmetry can often simplify the problem by reducing the number of components to consider.

Answer 1

The Correct Option is C

Step 1: Electric field due to a point charge.
The electric field at the origin due to a point charge is given by: \[ E = \frac{kq}{r^2} \] where \( k = \frac{1}{4\pi \epsilon_0} \) is Coulomb’s constant. Step 2: Calculate the electric field due to multiple charges.
For the given configuration, the fields from the charges will combine vectorially. Since the charges are symmetrically placed on a circle, the components of the electric field along the radial directions cancel each other, and only the components along the \( x \)-axis remain. Step 3: Conclusion.
After summing the contributions from all the charges, we get the resultant electric field as \( \frac{\sqrt{3q}}{2\pi \epsilon_0 r^2} (-\hat{i}) \). Final Answer: \[ \boxed{\frac{\sqrt{3q}}{2\pi \epsilon_0 r^2} (-\hat{i})} \]