Two charges 2*10^-6 and 1*10^-6 are placed at a separation of 10 cm. Where should a third charge be placed such that it experience no net force due to these charges?

To solve this problem, we need to determine the position where a third charge experiences no net force due to the given charges.

### Given data:
- Charge 1: \( q_1 = 2 \times 10^{-6} \) C
- Charge 2: \( q_2 = 1 \times 10^{-6} \) C
- Distance between charges: \( d = 10 \) cm = 0.1 m

Let the third charge be \( q_3 \), and let it be placed at a position \( x \) from \( q_1 \) along the line joining the two charges. There are two possible regions where \( q_3 \) could be placed:
1. **To the left of \( q_1 \)**
2. **Between \( q_1 \) and \( q_2 \)**
3. **To the right of \( q_2 \)**

We will examine each case.

### Case 1: Placing \( q_3 \) between \( q_1 \) and \( q_2 \)

Let \( q_3 \) be at a distance \( x \) from \( q_1 \), so its distance from \( q_2 \) is \( (0.1 - x) \). The forces due to both charges should cancel each other.

Using Coulomb's law, the force on \( q_3 \) due to \( q_1 \) is:

\[
F_1 = \frac{k q_1 q_3}{x^2}
\]

The force on \( q_3 \) due to \( q_2 \) is:

\[
F_2 = \frac{k q_2 q_3}{(0.1 - x)^2}
\]

For equilibrium, \( F_1 = F_2 \):

\[
\frac{k q_1 q_3}{x^2} = \frac{k q_2 q_3}{(0.1 - x)^2}
\]

Cancel \( k \) and \( q_3 \) from both sides:

\[
\frac{q_1}{x^2} = \frac{q_2}{(0.1 - x)^2}
\]

Substituting values:

\[
\frac{2 \times 10^{-6}}{x^2} = \frac{1 \times 10^{-6}}{(0.1 - x)^2}
\]

Dividing both sides by \( 10^{-6} \):

\[
\frac{2}{x^2} = \frac{1}{(0.1 - x)^2}
\]

Cross multiplying:

\[
2(0.1 - x)^2 = x^2
\]

Expanding:

\[
2(0.01 - 0.2x + x^2) = x^2
\]

\[
0.02 - 0.4x + 2x^2 = x^2
\]

Rearrange:

\[
2x^2 - x^2 - 0.4x + 0.02 = 0
\]

\[
x^2 - 0.4x + 0.02 = 0
\]

Solving this quadratic equation using the quadratic formula:

\[
x = \frac{-(-0.4) \pm \sqrt{(-0.4)^2 - 4(1)(0.02)}}{2(1)}
\]

\[
x = \frac{0.4 \pm \sqrt{0.16 - 0.08}}{2}
\]

\[
x = \frac{0.4 \pm \sqrt{0.08}}{2}
\]

Approximating \( \sqrt{0.08} \approx 0.282 \):

\[
x = \frac{0.4 \pm 0.282}{2}
\]

Solving for two values:

\[
x_1 = \frac{0.4 + 0.282}{2} = \frac{0.682}{2} = 0.341
\]

\[
x_2 = \frac{0.4 - 0.282}{2} = \frac{0.118}{2} = 0.059
\]

Since \( x = 0.341 \) m is not between \( q_1 \) and \( q_2 \), we reject it. The valid solution is:

\[
x = 0.059 \text{ m} = 5.9 \text{ cm from } q_1
\]

### Conclusion:
The third charge should be placed **5.9 cm from the charge \( q_1 \) towards \( q_2 \)** along the line joining the two charges to experience zero net force.