Question

In the \( (x, y, z) \) coordinate system, three point charges \( Q \), \( Q \), and \( \alpha Q \) are located in free space at \( (-1, 0, 0) \), \( (1, 0, 0) \), and \( (0, -1, 0) \), respectively. The value of \( \alpha \) for the electric field to be zero at \( (0, 0.5, 0) \) is \_\_\_\_\_\_ (rounded off to 1 decimal place).

Hint:

When solving for the electric field, ensure the superposition principle is applied correctly. Calculate contributions from each charge and resolve components along the axes for symmetry.

Answer 1

Step 1: Determine the electric field at \( (0, 0.5, 0) \) due to charges at \( (-1, 0, 0) \) and \( (1, 0, 0) \). The electric field at \( (0, 0.5, 0) \) due to a point charge \( Q \) is given by: \[ E = \frac{k Q}{r^2} \hat{r} \] where \( k = \frac{1}{4 \pi \epsilon_0} \) and \( r \) is the distance between the charge and the point. The contributions from the charges at \( (-1, 0, 0) \) and \( (1, 0, 0) \) are symmetric, resulting in a combined field along the \( y \)-direction: \[ E_{net} = 2E_y = 2 \cdot \frac{k Q}{(1^2 + 0.5^2)} \cdot \frac{0.5}{\sqrt{1^2 + 0.5^2}} \] Simplifying: \[ E_{net} = 2 \cdot \frac{k Q}{1.25} \cdot 0.357 \approx 0.715 k Q \, \hat{y}. \] Step 2: Add the contribution from the charge at \( (0, -1, 0) \). The electric field due to \( \alpha Q \) at \( (0, 0.5, 0) \) is: \[ E_{\alpha Q} = \frac{k (\alpha Q)}{(1.5)^2} \, \hat{y} = \frac{0.444 k (\alpha Q)}{1} \, \hat{y}. \] Step 3: Set the total electric field to zero at \( (0, 0.5, 0) \). Combining all contributions: \[ E_{net} = 0.715 k Q \, \hat{y} + 0.444 k (\alpha Q) \, \hat{y} = 0. \] Simplify: \[ 0.715 Q + 0.444 (\alpha Q) = 0. \] Dividing by \( k Q \): \[ 0.715 + 0.444 \alpha = 0. \] Solve for \( \alpha \): \[ \alpha = \frac{-0.715}{0.444} \approx -1.61. \]