JEE Physics 2026 Question with Solution

Given: Six point charges are placed on the circumference of a circle of radius $R$ at angular separation $60^\circ$.

Find: The net electric field at the center.

Each charge is placed at a distance $R$ from the center. The magnitude of electric field due to a charge $Q$ at the center is

$$E = \frac{1}{4\pi \varepsilon_0}\frac{Q}{R^2}$$

Step 1: Analyze symmetry of charge distribution.

From the figure, the charges are placed at angular separations of $60^\circ$. Due to symmetry, electric field vectors due to opposite charges partially cancel each other. Hence, we resolve the electric field vectors along the $x$-axis and $y$-axis.

Step 2: Resolve horizontal components.

Considering the directions of electric field due to each charge, the net horizontal component is proportional to

$$-\sqrt{3}$$

Step 3: Resolve vertical components.

Similarly, the net vertical component is proportional to

$$+1$$

Step 4: Write the resultant electric field vector.

Combining both components,

$$\vec{E} = -\frac{Q}{4\pi \varepsilon_0 R^2}\left(\sqrt{3}\,\hat{i}-\hat{j}\right)$$

Therefore, the correct option is B.

Given: The electric field at the center is to be obtained by vector addition of fields due to all six charges.

Find: The resultant in unit-vector form.

Use the fact that each contribution has the same magnitude

$$\frac{Q}{4\pi \varepsilon_0 R^2}$$

but different direction depending on the position and sign of the charge.

The solution indicates that after resolving all fields:

$$E_x \propto -\sqrt{3}, \qquad E_y \propto +1$$

So the net field is

$$\vec{E} = \frac{Q}{4\pi \varepsilon_0 R^2}\left(-\sqrt{3}\,\hat{i}+\hat{j}\right)$$

which is equivalently written as

$$\vec{E} = -\frac{Q}{4\pi \varepsilon_0 R^2}\left(\sqrt{3}\,\hat{i}-\hat{j}\right)$$

Hence, the final answer is $-\dfrac{Q}{4\pi \varepsilon_0 R^2}\left(\sqrt{3}\,\hat{i}-\hat{j}\right)$, so the correct option is B.