1. Given charges and point of interest

• Charge $q_1 = 10\,\mu C$ is at the origin, i.e. at $x=0$.
• Charge $q_2 = 40\,\mu C$ is to be placed somewhere on the $x$-axis.
• Net electric field must be zero at the point $x=2\,\text{cm}$.

Since both charges are positive, the electric field at a point is directed away from each charge.

2. Electric field due to $q_1$ at $x=2\,\text{cm}$

Point $x=2\,\text{cm}$ lies to the right of the origin, so the field due to $q_1=10\,\mu C$ is toward the right.

Its magnitude is

where distances are in cm consistently for comparison.

3. For net field to be zero

The field due to $q_2=40\,\mu C$ at $x=2\,\text{cm}$ must be equal in magnitude and opposite in direction.

So $q_2$ must be placed to the right of $x=2$ cm, so that its field at $x=2$ cm points to the left.

Let its position be $x=a$ cm, with $a>2$.

Then distance from $q_2$ to the point $x=2$ cm is

Hence,

For cancellation,

So,

Cancel $k$:

Thus,

4. Check direction condition

We need the field due to $q_2$ at $x=2$ cm to point left, opposite to $E_1$.

• If $a=6$ cm, then point $x=2$ cm is to the left of $q_2$, so field due to positive $q_2$ points left. Correct.
• If $a=-2$ cm, then point $x=2$ cm is to the right of $q_2$, so field due to positive $q_2$ points right. This would not cancel $E_1$.

So the only valid position is

5. Option check

• A: $x=6$ cm ✅
• B: $x=8$ cm ❌
• C: $x=4$ cm ❌
• D: $x=-4$ cm ❌

Therefore, the correct option is A.