Let's break down the problem **step by step**. --- ### **Given Data** - Two charges on x-axis: - \( q_1 = -1.60 \times 10^{-9} \) C at \( x_1 = -3.50 \) m - \( q_2 = +1.60 \times 10^{-9} \) C at \( x_2 = +3.50 \) m - Point \( P \) at \( (0, 5.00) \) m (on the y-axis) - Find **x and y components** of the net electric field at \( P \). - Coulomb's constant: \( k = 8.99 \times 10^9 \ \mathrm{N \cdot m^2 / C^2} \) --- ## **Step 1: Calculate Distance from Each Charge to P** For both charges, the distance to \( P \) is: \[ r = \sqrt{(x_p - x_q)^2 + (y_p - y_q)^2} \] - For both charges, \( y_q = 0 \), \( y_p = 5.00 \), and \( x_p = 0 \). - For \( q_1 \) at \( x_1 = -3.50 \): \[ r_1 = \sqrt{(-3.50 - 0)^2 + (0 - 5.00)^2} = \sqrt{3.5^2 + 5^2} = \sqrt{12.25 + 25} = \sqrt{37.25} \approx 6.105 \text{ m} \] - For \( q_2 \) at \( x_2 = +3.50 \): \[ r_2 = \sqrt{(3.50 - 0)^2 + (0 - 5.00)^2} = \sqrt{3.5^2 + 5^2} = \sqrt{37.25} \approx 6.105 \text{ m} \] - Both distances are the same. --- ## **Step 2: Find Direction Cosines for Each Charge** For both charges: - \( \Delta x = x_q - x_p = \pm 3.5 - 0 \) - \( \Delta y = y_q - y_p = 0 - 5 = -5 \) The unit vector from each charge to P: \[ \hat{r} = \frac{(\mp 3.5)\hat{i} + (-5)\hat{j}}{6.105} \] --- ## **Step 3: Electric Field Due to a Point Charge** Magnitude: \[ E = \frac{k |q|}{r^2} \] For both charges: \[ E_1 = E_2 = \frac{8.99 \times 10^9 \times 1.60 \times 10^{-9}}{(6.105)^2} \] \[ = \frac{8.99 \times 1.60 \times 10^0}{37.25} \] \[ = \frac{14.384}{37.25} \approx 0.386 \ \mathrm{N/C} \] --- ## **Step 4: Components for Each Charge** ### **For \( q_1 \) (at \( x_1 = -3.5 \), negative charge)** The electric field points **toward** the negative charge, i.e., from P to the charge: \[ \vec{E}_1 = E_1 \cdot \frac{(-3.5)\hat{i} + (-5)\hat{j}}{6.105} \] - \( E_{1x} = 0.386 \times \frac{-3.5}{6.105} = 0.386 \times -0.5734 = -0.221 \ \mathrm{N/C} \) - \( E_{1y} = 0.386 \times \frac{-5}{6.105} = 0.386 \times -0.8187 = -0.316 \ \mathrm{N/C} \) --- ### **For \( q_2 \) (at \( x_2 = +3.5 \), positive charge)** The electric field points **away** from the positive charge, i.e., from the charge to P: \[ \vec{E}_2 = E_2 \cdot \frac{(0 - 3.5)\hat{i} + (5 - 0)\hat{j}}{6.105} = E_2 \cdot \frac{-3.5\hat{i} + 5\hat{j}}{6.105} \] - \( E_{2x} = 0.386 \times \frac{-3.5}{6.105} = -0.221 \ \mathrm{N/C} \) - \( E_{2y} = 0.386 \times \frac{5}{6.105} = +0.316 \ \mathrm{N/C} \) --- ## **Step 5: Add Components** ### **x-components:** \[ E_x = E_{1x} + E_{2x} = (-0.221) + (-0.221) = -0.442 \ \mathrm{N/C} \] ### **y-components:** \[ E_y = E_{1y} + E_{2y} = (-0.316) + 0.316 = 0 \] --- ## **Final Answers** ### **a) \( x \)-component:** \[ \boxed{-0.442 \ \mathrm{N/C}} \] ### **b) \( y \)-component:** \[ \boxed{0 \ \mathrm{N/C}} \] --- ### **Summary Table** | Component | Value | |---------------|---------------------| | \(E_x\) | \(-0.442\) N/C | | \(E_y\) | \(0\) N/C | --- **That's the step-by-step solution! If you need further explanation or calculations, let me know!**

The net electric field at a point is obtained by vectorially adding the contributions from all individual point charges, considering both magnitude and direction. Components along axes are calculated using Coulomb's law and unit vectors, then summed to find the resultant field.