A point charge of 10 $$\mu$$C is placed at the origin. At what location on the X-axis should a point charge of 40 $$\mu$$C be placed so that the net electric field is zero at $$x = 2$$ cm on the X-axis?

At point $$x = 2$$ cm, the net electric field must be zero.

Charge at origin: $$q_1 = 10$$ μC (positive). At $$x = 2$$ cm, the field due to $$q_1$$ points in the +x direction.

For the net field to be zero, the 40 μC charge must be placed to the right of $$x = 2$$ (so its field at x=2 points in the -x direction).

Let the 40 μC charge be at position $$x = d$$ cm where $$d > 2$$.

At $$x = 2$$:

$$\frac{kq_1}{(2)^2} = \frac{kq_2}{(d-2)^2}$$

$$\frac{10}{4} = \frac{40}{(d-2)^2}$$

$$(d-2)^2 = \frac{40 \times 4}{10} = 16$$

$$d - 2 = 4$$

$$d = 6$$ cm

The correct answer is Option 1: $$x = 6$$ cm.