To determine the position \(x_0\) for the second charge (\(40 \, \mu \text{C}\)) where the net electric field is zero at \(x = 2 \, \text{cm}\): Step 1: Equate Electric Field Magnitudes The net electric field is zero when the magnitudes of the fields from both charges are equal: \[K \cdot \frac{10 \times 10^{-6}}{(2)^2} = K \cdot \frac{40 \times 10^{-6}}{(x_0 - 2)^2}.\] Step 2: Simplify the Expression Cancel \(K\) and simplify the numerical values: \[\frac{10}{4} = \frac{40}{(x_0 - 2)^2}.\] Further simplification yields: \[\frac{1}{2} = \frac{2}{(x_0 - 2)^2}.\] Step 3: Solve for \(x_0\) Rearrange the equation to isolate \( (x_0 - 2)^2 \): \[(x_0 - 2)^2 = 4.\] Taking the square root of both sides gives: \[x_0 - 2 = \pm 2.\] The possible values for \(x_0\) are: \[x_0 = 4 \, \text{cm} \quad \text{or} \quad x_0 = 6 \, \text{cm}.\] Step 4: Select the Valid Position Given that the second charge is located to the right of \(x = 2 \, \text{cm}\), the correct position is: \[x_0 = 6 \, \text{cm}.\] Final Answer: \[\boxed{x_0 = 6 \, \text{cm}}\]