# Step-by-Step Solution: Electric Field at Point A Due to Three Charges Let’s solve the problem step by step. We are given three charges: - \( Q_1 = 1\ \mu\text{C} \) - \( Q_2 = 10\ \mu\text{C} \) - \( Q_3 = -3\ \mu\text{C} \) The arrangement is not specified, but let's assume a typical configuration for such questions: Suppose the charges are placed at the corners of a right triangle, and you are to find the field at point \( A \), one of these corners. For a complete answer, let’s define coordinates: - Place \( Q_1 = 1\ \mu\text{C} \) at the origin, \( (, ) \). - Place \( Q_2 = 10\ \mu\text{C} \) at \( (a, ) \). - Place \( Q_3 = -3\ \mu\text{C} \) at \( (, a) \). - Point \( A \) is at the origin. Let’s find the electric field at \( A = (, ) \) from the other two charges. ## 1. Electric Field Formula For a point charge: \[ \vec{E} = \frac{1}{4\pi\epsilon_} \frac{Q}{r^2} \hat{r} \] where - \( Q \) is the charge, - \( r \) is the distance from the charge to the point, - \( \hat{r} \) is the unit vector from the charge to the point. Let \( k = \frac{1}{4\pi\epsilon_} = 8.99 \times 10^9 \ \text{N m}^2/\text{C}^2 \). ## 2. Find Electric Field at \( A \) from Each Charge ### (a) Contribution from \( Q_2 \) - \( Q_2 = 10\ \mu\text{C} = 10 \times 10^{-6}\ \text{C} \) - Position: \( (a, ) \) → distance to \( A \): \( |a| \) - Direction: from \( Q_2 \) to \( A \) (negative x-direction) \[ E_{2} = k \frac{|Q_2|}{a^2} \] - Since \( Q_2 \) is positive, field at \( A \) points **away** from \( Q_2 \), i.e., towards negative x-direction: \( (-1, ) \). \[ \vec{E}_2 = -k \frac{Q_2}{a^2} \hat{i} \] ### (b) Contribution from \( Q_3 \) - \( Q_3 = -3\ \mu\text{C} = -3 \times 10^{-6}\ \text{C} \) - Position: \( (, a) \) → distance to \( A \): \( |a| \) - Direction: from \( Q_3 \) to \( A \) (negative y-direction) \[ E_{3} = k \frac{|Q_3|}{a^2} \] - Since \( Q_3 \) is negative, field at \( A \) points **toward** \( Q_3 \), i.e., towards negative y-direction: \( (, -1) \). \[ \vec{E}_3 = -k \frac{Q_3}{a^2} \hat{j} \] Note the double negative: \( -(-3 \times 10^{-6}) = 3 \times 10^{-6} \), so the field is in the negative y-direction. ### (c) Contribution from \( Q_1 \) - \( Q_1 \) is at \( A \), so its field at \( A \) is **undefined/infinite** (a charge exerts no field on itself). ## 3. Total Electric Field at \( A \) \[ \vec{E}_A = \vec{E}_2 + \vec{E}_3 = \left( -k \frac{Q_2}{a^2} \right) \hat{i} + \left( -k \frac{Q_3}{a^2} \right) \hat{j} \] Plug in values: - \( Q_2 = 10 \times 10^{-6}\ \text{C} \) - \( Q_3 = -3 \times 10^{-6}\ \text{C} \) - \( k = 8.99 \times 10^9 \) - \( a \): not given, so leave as variable. \[ \vec{E}_A = -8.99 \times 10^9 \frac{10 \times 10^{-6}}{a^2} \hat{i} -8.99 \times 10^9 \frac{-3 \times 10^{-6}}{a^2} \hat{j} \] \[ = (-8.99 \times 10^4/a^2) \hat{i} + (2.697 \times 10^4/a^2) \hat{j} \] Units: N/C ## 4. Final Vector Answer (in n/C) 1 N/C = \( 10^9 \) n/C \[ \vec{E}_A = \left(-8.99 \times 10^4/a^2 \right) \hat{i} + \left(2.697 \times 10^4/a^2 \right) \hat{j} \ \text{N/C} \] \[ \vec{E}_A = \left(-8.99 \times 10^{13}/a^2\right) \hat{i} + \left(2.697 \times 10^{13}/a^2\right) \hat{j} \ \text{n/C} \] --- ## **Summary** **The electric field at point \( A \) due to all three charges is:** \[ \boxed{ \vec{E}_A = \left(-8.99 \times 10^{13}/a^2\right) \hat{i} + \left(2.697 \times 10^{13}/a^2\right) \hat{j} \quad \text{n/C} } \] where \( a \) is the distance (in meters) from \( A \) to each of the other two charges. **Direction:** - The field points left (negative x) and up (positive y). --- **Note:** If you are given a specific value for \( a \) or a different arrangement, substitute accordingly. If you can upload the figure or clarify positions, I can tailor the answer exactly.