# Electric Field at the Center of an Equilateral Triangle of Charges We have **three point charges** at the corners of an equilateral triangle (side \( s = 22.\, \text{cm} = .220\, \text{m} \)): - \( q_a = 2.30\, \text{nC}) - \( q_b = -7.50\, \text{nC} \) - \( q_c = 1.50\, \text{nC} \) Find the **electric field at the center** (magnitude and direction wrt the \( +x \)-axis). --- ## 1. **Geometry & Center Coordinates** Let’s place the triangle in the \( xy \)-plane for calculation: - Vertex A at \( (, s/\sqrt{3}) \) - Vertex B at \( (-s/2, -s/(2\sqrt{3})) \) - Vertex C at \( (s/2, -s/(2\sqrt{3})) \) The **center** (centroid) is at the origin \( (,) \). The **distance from center to each vertex** is: \[ r = \frac{s}{\sqrt{3}} = \frac{.220}{1.732} = .127\, \text{m} \] --- ## 2. **Electric Field Due to Each Charge** The electric field from a point charge at distance \( r \): \[ E = \frac{1}{4\pi\epsilon_} \frac{|q|}{r^2} \] where \( \frac{1}{4\pi\epsilon_} = 8.99 \times 10^9\, \text{N}\cdot\text{m}^2/\text{C}^2 \). --- ### **A. Coordinates and Directions** - \( q_a \) at \( (, r) \) (top) - \( q_b \) at \( (-s/2, -r/2) \) - \( q_c \) at \( (s/2, -r/2) \) Find the **unit vector from each charge to the center** (since the field at the center is due to the charge at the vertex, pointing from the charge to the center): #### - For \( q_a \): - Vector from \( (, r) \) to \( (, ) \): \( \vec{r}_a = (, -r) \) - Unit vector: \( \hat{r}_a = (, -1) \) #### - For \( q_b \): - Vertex: \( (-s/2, -r/2) \) - Vector to center: \( (s/2, r/2) \) - Magnitude: \( \sqrt{ (s/2)^2 + (r/2)^2 } = r \), since \( r = s/\sqrt{3} \) and can be shown by calculation. - Unit vector: \( \hat{r}_b = \left( \frac{s}{2r}, \frac{1}{2} \right) \) #### - For \( q_c \): - Vertex: \( (s/2, -r/2) \) - Vector to center: \( (-s/2, r/2) \) - Unit vector: \( \hat{r}_c = \left( -\frac{s}{2r}, \frac{1}{2} \right) \) With \( \frac{s}{2r} = \frac{.220}{2 \times .127} = .866 \). --- ### **B. Calculate Each Electric Field** \[ E_i = \frac{k |q_i|}{r^2} \] where \( k = 8.99 \times 10^9 \), \( r = .127\, \text{m} \). #### - For \( q_a = 2.30\, \text{nC} \): \[ E_a = \frac{8.99 \times 10^9 \times 2.30 \times 10^{-9}}{(.127)^2} = \frac{20.677 \times 10^}{.016129} = 1282\, \text{N/C} \] - Direction: \( (, -1) \) #### - For \( q_b = -7.50\, \text{nC} \): \[ E_b = \frac{8.99 \times 10^9 \times 7.50 \times 10^{-9}}{(.127)^2} = \frac{67.425}{.016129} = 4183\, \text{N/C} \] - Since \( q_b \) is negative, field is toward \( q_b \), i.e., **opposite** the direction from \( q_b \) to center \( \Rightarrow \) direction \( (-.866, -.5) \). #### - For \( q_c = 1.50\, \text{nC} \): \[ E_c = \frac{8.99 \times 10^9 \times 1.50 \times 10^{-9}}{(.127)^2} = \frac{13.485}{.016129} = 836\, \text{N/C} \] - Direction: \( (-.866, .5) \) --- ## 3. **Express All Vectors in Components** #### - \( \vec{E}_a \): \[ \vec{E}_a = 1282 \times (, -1) = (, -1282) \] #### - \( \vec{E}_b \) (negative charge, field toward the charge): \[ \vec{E}_b = 4183 \times (-.866, -.5) = (-3622, -2092) \] #### - \( \vec{E}_c \) (positive charge, field away from charge): \[ \vec{E}_c = 836 \times (-.866, .5) = (-724, 418) \] --- ## 4. **Sum All Components** \[ E_x = + (-3622) + (-724) = -4346\, \text{N/C} \] \[ E_y = -1282 + (-2092) + 418 = -2956\, \text{N/C} \] --- ## 5. **Find Magnitude and Direction** ### **A. Magnitude** \[ E = \sqrt{E_x^2 + E_y^2} = \sqrt{(-4346)^2 + (-2956)^2} = \sqrt{18889116 + 8736716} = \sqrt{27625832} = 5256\, \text{N/C} \] ### **B. Direction (wrt +x-axis)** \[ \theta = \arctan\left( \frac{E_y}{E_x} \right) = \arctan\left( \frac{-2956}{-4346} \right) = \arctan(.680) = 34.2^\circ \] Both \( E_x \) and \( E_y \) are negative, so the field points in the **third quadrant**. \[ \text{Angle from +x-axis} = 180^\circ + 34.2^\circ = 214.2^\circ \] --- ## **Final Answers** **Magnitude:** \[ \boxed{E = 5.26 \times 10^3\, \text{N/C}} \] **Direction (from +x-axis):** \[ \boxed{214^\circ} \] --- ## **Summary Table** | Charge | Value (nC) | Position | \( E \) (N/C) | Direction (unit vector) | Components (N/C) | |--------|------------|---------------------|---------------|-------------------------------|--------------------------| | \(q_a\)| +2.30 | (, +r) | 1282 | (, -1) | (, -1282) | | \(q_b\)| -7.50 | (-s/2, -r/2) | 4183 | (-.866, -.5) | (-3622, -2092) | | \(q_c\)| +1.50 | (s/2, -r/2) | 836 | (-.866, .5) | (-724, 418) | | **Sum**| — | — | **5256** | — | (-4346, -2956) | --- **Final Result:** **Magnitude** \( E = 5.26 \times 10^3\, \text{N/C} \) **Direction** \( = 214^\circ \) (counterclockwise from \( +x \)-axis)

# Electric Field at the Center of a Triangular Configuration of Charges We need to find the electric field's magnitude and direction at the center of an equilateral triangle formed by three charges: - \( q_a = 2.30\, \text{nC} \) - \( q_b = -7.50\, \text{nC} \) - \( q_c = 1.50\, \text{nC} \) The triangle has a side length of \( s = 22.0\, \text{cm} = 0.220\, \text{m} \). --- ## 1. **Geometry and Center Coordinates** ### Triangle Placement - Place vertex A at \( (0, \frac{s}{\sqrt{3}}) \) - Vertex B at \( (-\frac{s}{2}, -\frac{s}{2\sqrt{3}}) \) - Vertex C at \( (\frac{s}{2}, -\frac{s}{2\sqrt{3}}) \) ### Center (Centroid) The centroid is located at the origin \( (0, 0) \). The distance from the center to each vertex is given by: \[ r = \frac{s}{\sqrt{3}} = \frac{0.220}{\sqrt{3}} \approx 0.127\, \text{m} \] --- ## 2. **Electric Field Due to Each Charge** The electric field \( E \) due to a point charge is given by: \[ E = \frac{1}{4\pi\epsilon_0} \frac{|q|}{r^2} \] where \( \frac{1}{4\pi\epsilon_0} = 8.99 \times 10^9\, \text{N}\cdot\text{m}^2/\text{C}^2 \). --- ### 2.1 **Unit Vectors from Each Charge to the Center** - **For \( q_a \):** - Direction towards center: \( (0, -1) \) - **For \( q_b \):** - Direction towards center: \( \left( \frac{\frac{s}{2}}{r}, \frac{\frac{s}{2\sqrt{3}}}{r} \right) = \left( \frac{.220/2}{.127}, \frac{-.220/(2\sqrt{3})}{.127} \right) \approx (0.866, -0.5) \) - **For \( q_c \):** - Direction towards center: \( \left( -\frac{\frac{s}{2}}{r}, \frac{\frac{s}{2\sqrt{3}}}{r} \right) = \left( -0.866, -0.5 \right) \) --- ## 3. **Calculate Each Electric Field** ### 3.1 **Electric Fields from Each Charge** - **For \( q_a = 2.30\, \text{nC} \):** \[ E_a = \frac{8.99 \times 10^9 \times 2.30 \times 10^{-9}}{(0.127)^2} \approx 1282\, \text{N/C} \] - **For \( q_b = -7.50\, \text{nC} \):** \[ E_b = \frac{8.99 \times 10^9 \times 7.50 \times 10^{-9}}{(0.127)^2} \approx 4183\, \text{N/C} \] - **For \( q_c = 1.50\, \text{nC} \):** \[ E_c = \frac{8.99 \times 10^9 \times 1.50 \times 10^{-9}}{(0.127)^2} \approx 836\, \text{N/C} \] --- ## 4. **Express All Vectors in Components** ### 4.1 **Components of Each Electric Field** - **\( \vec{E}_a \):** \[ \vec{E}_a = 1282 \times (0, -1) = (0, -1282) \] - **\( \vec{E}_b \):** \[ \vec{E}_b = 4183 \times (0.866, -0.5) = (3622.78, -2091.5) \] - **\( \vec{E}_c \):** \[ \vec{E}_c = 836 \times (-0.866, 0.5) = (-724.176, 418) \] --- ## 5. **Sum All Components** ### 5.1 **Resultant Electric Field Components** \[ E_x = 0 + 3622.78 - 724.176 \approx 2898.60\, \text{N/C} \] \[ E_y = -1282 - 2091.5 + 418 \approx -2955.50\, \text{N/C} \] --- ## 6. **Find Magnitude and Direction** ### 6.1 **Magnitude of the Electric Field** \[ E = \sqrt{E_x^2 + E_y^2} = \sqrt{(2898.60)^2 + (-2955.50)^2} \approx 4163.38\, \text{N/C} \] ### 6.2 **Direction (Angle from +x-axis)** \[ \theta = \tan^{-1}\left( \frac{E_y}{E_x} \right) = \tan^{-1}\left( \frac{-2955.50}{2898.60} \right) \approx -45.0^\circ \] Since the angle is negative, it indicates an angle measured clockwise from the positive x-axis. --- ## **Final Answers** **Magnitude:** \[ \boxed{E \approx 4.16 \times 10^3\, \text{N/C}} \] **Direction (from +x-axis):** \[ \boxed{-45^\circ} \] This indicates the electric field is directed downward and to the left, in the fourth quadrant.