Question

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 \( x = 2 \, \text{cm} \) on the X-axis?

• A. \( 6 \, \text{cm} \)
• B. \( 4 \, \text{cm} \)
• C. \( 8 \, \text{cm} \)
• D. \( -4 \, \text{cm} \)

Hint:

When determining the position for zero net electric field, equate the magnitudes of electric fields from the charges and carefully solve the resulting equation. Consider the geometry and direction of forces to select the correct solution.

Answer 1

The Correct Option is A

To find the position \(x_0\) where the second charge (\(40 \, \mu \text{C}\)) should be located so that the net electric field is zero at \(x = 2 \, \text{cm}\): Step 1: Balance the Electric Fields The net electric field at \(x = 2 \, \text{cm}\) is zero when the contributions from both charges cancel out: \[ E_{\text{net}} = 0 \implies K \cdot \frac{10 \times 10^{-6}}{(2)^2} = K \cdot \frac{40 \times 10^{-6}}{(x_0 - 2)^2}. \] Step 2: Simplify the Equation Divide out \(K\) and simplify the terms: \[ \frac{10}{4} = \frac{40}{(x_0 - 2)^2}. \] \[ \frac{1}{2} = \frac{2}{(x_0 - 2)^2}. \] Step 3: Solve for \(x_0\) Rearrange the equation: \[ (x_0 - 2)^2 = 4. \] Take the square root of both sides: \[ x_0 - 2 = \pm 2. \] Thus: \[ x_0 = 4 \, \text{cm} \quad \text{or} \quad x_0 = 6 \, \text{cm}. \] Step 4: Choose the Correct Value of \(x_0\) Since the second charge is positioned to the right of \(x = 2 \, \text{cm}\), the correct location is: \[ x_0 = 6 \, \text{cm}. \] Final Answer: \[ \boxed{x_0 = 6 \, \text{cm}} \]