Let's break down the problem step by step: ## **Step 1: Understanding the Setup** - We have 4 charges, each of magnitude \( q = e = 1.6 \times 10^{-19} \) C (the charge of an electron). - The charges are arranged on a circle of radius \( r = 0.04 \) m, at angles \( 15^\circ, 120^\circ, 225^\circ, 300^\circ \). - We are to find the net electric field at the origin (center of the circle). --- ## **Step 2: Electric Field Due to a Single Point Charge** The electric field due to a point charge at a distance \( r \) is: \[ E = \frac{k|q|}{r^2} \] where \( k = 8.99 \times 10^9 \ \mathrm{N\cdot m^2/C^2} \). --- ## **Step 3: Direction of Each Electric Field** Each field at the origin points **away from positive charge** (or toward if negative). Assume all are positive for now. The direction of the electric field at the origin due to each charge is **directly opposite** to the position vector of the charge. --- ## **Step 4: Components of Each Field** Let’s calculate the field vector for each charge. Let \( E_0 = \frac{k|q|}{r^2} \). - \( E_1 \): At \( 15^\circ \rightarrow \) Field at \( 15^\circ + 180^\circ = 195^\circ \) - \( E_2 \): At \( 120^\circ \rightarrow \) Field at \( 300^\circ \) - \( E_3 \): At \( 225^\circ \rightarrow \) Field at \( 45^\circ \) - \( E_4 \): At \( 300^\circ \rightarrow \) Field at \( 120^\circ \) --- ### **Field Components** For each field, \[ \vec{E}_i = E_0 (\cos\theta_i\ \hat{i} + \sin\theta_i\ \hat{j}) \] where \( \theta_i \) is the angle of the field direction. So, - \( \theta_1 = 195^\circ \) - \( \theta_2 = 300^\circ \) - \( \theta_3 = 45^\circ \) - \( \theta_4 = 120^\circ \) --- ## **Step 5: Calculate \( E_0 \)** \[ E_0 = \frac{8.99 \times 10^9 \times 1.6 \times 10^{-19}}{(0.04)^2} \] \[ = \frac{8.99 \times 10^9 \times 1.6 \times 10^{-19}}{0.0016} \] \[ = \frac{1.4384 \times 10^{-9}}{0.0016} \] \[ = 8.99 \times 10^9 \times \frac{1.6 \times 10^{-19}}{0.0016} = 8.99 \times 10^9 \times 1 \times 10^{-16} = 8.99 \times 10^{-7} \ \mathrm{N/C} \] --- ## **Step 6: Find the Total Field (Vector Sum)** Add the x and y components: \[ E_{x,\text{total}} = E_0 \left[\cos 195^\circ + \cos 300^\circ + \cos 45^\circ + \cos 120^\circ \right] \] \[ E_{y,\text{total}} = E_0 \left[\sin 195^\circ + \sin 300^\circ + \sin 45^\circ + \sin 120^\circ \right] \] Substitute: - \( \cos 195^\circ = -0.9659 \) - \( \cos 300^\circ = 0.5 \) - \( \cos 45^\circ = 0.7071 \) - \( \cos 120^\circ = -0.5 \) Sum: \[ E_{x,\text{total}} = E_0 [-0.9659 + 0.5 + 0.7071 - 0.5] = E_0 [-0.2588] \] - \( \sin 195^\circ = -0.2588 \) - \( \sin 300^\circ = -0.8660 \) - \( \sin 45^\circ = 0.7071 \) - \( \sin 120^\circ = 0.8660 \) Sum: \[ E_{y,\text{total}} = E_0 [-0.2588 - 0.8660 + 0.7071 + 0.8660] = E_0 [0.4483] \] --- ## **Step 7: Compute the Final Values** Plug in \( E_0 = 8.99 \times 10^{-7} \) N/C: \[ E_{x,\text{total}} = 8.99 \times 10^{-7} \times (-0.2588) = -2.33 \times 10^{-7} \ \mathrm{N/C} \] \[ E_{y,\text{total}} = 8.99 \times 10^{-7} \times 0.4483 = 4.03 \times 10^{-7} \ \mathrm{N/C} \] Magnitude: \[ E_{\text{total}} = \sqrt{(E_x)^2 + (E_y)^2} = \sqrt{(-2.33 \times 10^{-7})^2 + (4.03 \times 10^{-7})^2} = \sqrt{5.43 \times 10^{-14} + 1.63 \times 10^{-13}} = \sqrt{2.17 \times 10^{-13}} = 4.66 \times 10^{-7} \ \mathrm{N/C} \] Direction: \[ \theta = \tan^{-1}\left(\frac{E_{y,\text{total}}}{E_{x,\text{total}}}\right) = \tan^{-1}\left(\frac{4.03}{-2.33}\right) = \tan^{-1}(-1.73) \] This is in the **second quadrant**, so: \[ \theta = 180^\circ + \tan^{-1}(-1.73) = 180^\circ - 60^\circ = 120^\circ \] --- ## **Final Answer** ### **The net electric field at the origin has:** - **Magnitude:** \( \boxed{4.66 \times 10^{-7} \ \mathrm{N/C}} \) - **Direction:** \( \boxed{120^\circ} \) from the positive x-axis. --- If you have any further questions or need clarification, please let me know!