# Step-by-Step Solution: Electric Field at Point A Due to Three Charges Given: - \( Q_1 = 1\, \mu\mathrm{C} \) - \( Q_2 = 10\, \mu\mathrm{C} \) - \( Q_3 = -3\, \mu\mathrm{C} \) Let’s denote the positions: - Assume the positions and coordinates are as follows (since the figure is not provided, let's use a common layout for this kind of problem): | Charge | Position | |--------|---------------| | \(Q_1\) | (, ) | | \(Q_2\) | (a, ) | | \(Q_3\) | (, a) | | \(A\) | (a, a) | *You can adapt these coordinates if your figure is different. The process below applies to any such configuration.* ## 1. **Write the Electric Field Formula** The electric field due to a point charge is \[ \vec{E} = k \frac{q}{r^2} \hat{r} \] where \(k = 8.99 \times 10^9 \, \mathrm{N\,m^2/C^2}\), \(q\) is the charge, \(r\) is distance to the point of interest, and \(\hat{r}\) is the unit vector from the charge to the point. ## 2. **Find Vectors from Each Charge to Point A** - From \(Q_1\) at (, ) to \(A\) at (a, a): \(\vec{r}_A - \vec{r}_1 = (a, a)\) - From \(Q_2\) at (a, ) to \(A\) at (a, a): \(\vec{r}_A - \vec{r}_2 = (, a)\) - From \(Q_3\) at (, a) to \(A\) at (a, a): \(\vec{r}_A - \vec{r}_3 = (a, )\) ## 3. **Find Magnitudes and Unit Vectors** - \( |\vec{r}_A - \vec{r}_1| = \sqrt{a^2 + a^2} = a\sqrt{2} \) - \( |\vec{r}_A - \vec{r}_2| = a \) - \( |\vec{r}_A - \vec{r}_3| = a \) Unit vectors: - From \(Q_1\): \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \) - From \(Q_2\): \( (, 1) \) - From \(Q_3\): \( (1, ) \) ## 4. **Plug in the Charges (Convert to Coulombs)** - \( Q_1 = 1 \, \mu\mathrm{C} = 1 \times 10^{-6}\,\mathrm{C} \) - \( Q_2 = 10 \, \mu\mathrm{C} = 10 \times 10^{-6}\,\mathrm{C} \) - \( Q_3 = -3 \, \mu\mathrm{C} = -3 \times 10^{-6}\,\mathrm{C} \) ## 5. **Calculate the Electric Field from Each Charge at A** ### \( \vec{E}_1 \) (from \(Q_1\)): \[ \vec{E}_1 = k \frac{Q_1}{(a\sqrt{2})^2} \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) = k \frac{Q_1}{2a^2} \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \] ### \( \vec{E}_2 \) (from \(Q_2\)): \[ \vec{E}_2 = k \frac{Q_2}{a^2} (, 1) \] ### \( \vec{E}_3 \) (from \(Q_3\)): \[ \vec{E}_3 = k \frac{Q_3}{a^2} (1, ) \] ## 6. **Sum the Components** Let’s write all components: ### \( x \)-components: - \( \vec{E}_1: \) \( k \frac{Q_1}{2a^2} \frac{1}{\sqrt{2}} \) - \( \vec{E}_2: \) - \( \vec{E}_3: \) \( k \frac{Q_3}{a^2} \) ### \( y \)-components: - \( \vec{E}_1: \) \( k \frac{Q_1}{2a^2} \frac{1}{\sqrt{2}} \) - \( \vec{E}_2: \) \( k \frac{Q_2}{a^2} \) - \( \vec{E}_3: \) So, \[ E_x = k \frac{Q_1}{2a^2} \frac{1}{\sqrt{2}} + k \frac{Q_3}{a^2} \] \[ E_y = k \frac{Q_1}{2a^2} \frac{1}{\sqrt{2}} + k \frac{Q_2}{a^2} \] ## 7. **Plug in Values** Let’s choose \(a = 1\,\mathrm{m}\) (unless specified otherwise). - \(k = 8.99 \times 10^9\, \mathrm{N\,m^2/C^2}\) - \(\frac{1}{2\sqrt{2}} \approx .3536\) Plug in the charges: - \(Q_1 = 1 \times 10^{-6}\) - \(Q_2 = 10 \times 10^{-6}\) - \(Q_3 = -3 \times 10^{-6}\) ### Calculate Each Term: - \( k \frac{Q_1}{2a^2} \frac{1}{\sqrt{2}} = 8.99 \times 10^9 \times \frac{1 \times 10^{-6}}{2 \times 1^2} \times \frac{1}{\sqrt{2}} \) - \( \frac{1 \times 10^{-6}}{2} = .5 \times 10^{-6} \) - \( .5 \times 10^{-6} \times \frac{1}{\sqrt{2}} \approx .3536 \times 10^{-6} \) - \( 8.99 \times 10^9 \times .3536 \times 10^{-6} \approx 3178\, \mathrm{N/C} \) - \( k \frac{Q_2}{a^2} = 8.99 \times 10^9 \times 10 \times 10^{-6} = 8.99 \times 10^9 \times 10^{-5} = 8.99 \times 10^4 = 89,900\, \mathrm{N/C} \) - \( k \frac{Q_3}{a^2} = 8.99 \times 10^9 \times (-3) \times 10^{-6} = -26,970\, \mathrm{N/C} \) ### Add the components: #### \( x \)-component: \[ E_x = 3178 + (-26,970) = -23,792\, \mathrm{N/C} \] #### \( y \)-component: \[ E_y = 3178 + 89,900 = 93,078\, \mathrm{N/C} \] ## 8. **Express in Vector Form and Convert to n/C** 1 N/C = \(10^9\) n/C So, \[ E_x = -23,792\, \mathrm{N/C} = -2.38 \times 10^4\, \mathrm{N/C} = -2.38 \times 10^7\, \mathrm{n/C} \] \[ E_y = 93,078\, \mathrm{N/C} = 9.31 \times 10^4\, \mathrm{N/C} = 9.31 \times 10^7\, \mathrm{n/C} \] But as the problem likely wants the answer in n/C and the numbers are already manageable, you can keep it as: \[ \boxed{ \vec{E}_A = (-23,800\, \hat{x} + 93,100\, \hat{y})\, \mathrm{N/C} } \] or in n/C: \[ \boxed{ \vec{E}_A = (-23,800\, \hat{x} + 93,100\, \hat{y})\, \mathrm{n/C} } \] --- **Summary Table:** | Component | Value (N/C) | Value (n/C) | |-----------|------------------|------------------| | \(E_x\) | \(-23,800\) | \(-23,800\) | | \(E_y\) | \(93,100\) | \(93,100\) | --- ## **Final Answer** \[ \boxed{ \vec{E}_A = (-23,800\,\hat{x} + 93,100\,\hat{y})\, \mathrm{n/C} } \] --- **Note:** If your figure uses different coordinates or distances, adjust accordingly by plugging in the actual values for \(a\) and the vector directions. The **method** will be the same.