1. Given charges and positions

• $q_1 = 4q_0$ at $x=0$
• $q_2 = -q_0$ at $x=12\text{ cm}$
• A proton $(+q_0)$ is to be placed on the $x$-axis such that net electrostatic force on it is zero.

Since force on the proton is zero, the net electric field at its position must be zero.

2. Find possible region

Let the proton be at position $x$.

We check regions:

Region I: $x<0$

• Field due to $+4q_0$ points away from origin, i.e. toward left.
• Field due to $-q_0$ points toward $x=12$, i.e. toward right.

So fields are opposite; cancellation is possible.

Region II: $0<x<12$

• Field due to $+4q_0$ points to the right.
• Field due to $-q_0$ also points to the right (toward $x=12$).

Both fields are in same direction, so they cannot cancel.

Region III: $x>12$

• Field due to $+4q_0$ points to the right.
• Field due to $-q_0$ points to the left.

Fields are opposite; cancellation is possible.

3. Set magnitudes equal in the valid region

For cancellation,

for $x>12$.

Cancel $kq_0$:

Taking square root,

So,

4. Check the other algebraic root

From

we may also write

which gives

But $x=8$ lies between $0$ and $12$, where both fields are in the same direction, so this is not physically valid.

5. Check region $x<0$

If $x<0$, then distance from $q_1$ is $|x|=-x$ and from $q_2$ is $12-x$.

Equating magnitudes:

This gives no valid solution with $x<0$ after sign check, so no solution exists there.

6. Final answer

The proton must be placed at

from the origin.