Let's break down the problem step by step: ### **Step 1: Assign Positions and Charges** - \( q_1 > 0 \) at \((-4, 0)\) - \( q_2 = -\frac{1}{4}q_1 \) at \((-2, 0)\) - \( P \) at \((1, 0)\) - \( q_3 < 0 \) at \((1, 0)\) (same as point \(P\)) We need to find the electric field \(\vec{E}\) at point \(P\) due to \(q_1\) and \(q_2\), and then draw the force on \(q_3\). --- ### **Step 2: Electric Field Formula** The electric field at \(P\) due to a point charge \(q\) at position \((x_0, 0)\) is: \[ \vec{E} = k \frac{q}{r^2} \hat{r} \] where: - \(k = 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\) - \(r\) is the distance from the charge to \(P\) - \(\hat{r}\) is a unit vector from the charge to \(P\) --- ### **Step 3: Calculate Distances** For both charges, since all points are on the \(x\)-axis (\(y=0\)): - Distance from \(q_1\) to \(P\): \(r_1 = |1 - (-4)| = 5\) - Distance from \(q_2\) to \(P\): \(r_2 = |1 - (-2)| = 3\) --- ### **Step 4: Find Direction of \(\vec{E}\)** - Both charges are to the **left** of \(P\), so the field at \(P\) due to each charge is along the **positive \(x\)-axis** for positive charges, and **negative \(x\)-axis** for negative charges. --- ### **Step 5: Find Each Field** #### **a. Due to \(q_1\):** \[ \vec{E}_1 = k \frac{q_1}{(5)^2} \hat{i} \] Since \(q_1 > 0\), the field points away from \(q_1\), i.e., to the right (\(+\hat{i}\)). #### **b. Due to \(q_2\):** \[ q_2 = -\frac{1}{4} q_1 \] \[ \vec{E}_2 = k \frac{q_2}{(3)^2} \hat{i} \] Since \(q_2\) is negative, the field points **toward** \(q_2\), so still to the right (\(+\hat{i}\)), as \(P\) is to the right of \(q_2\). --- ### **Step 6: Add the Fields** Both fields are in the \(+\hat{i}\) direction: \[ \vec{E}_{\text{total}} = \vec{E}_1 + \vec{E}_2 \] \[ = k \frac{q_1}{25} \hat{i} + k \frac{-\frac{1}{4}q_1}{9} \hat{i} \] \[ = k q_1 \left(\frac{1}{25} - \frac{1}{36}\right) \hat{i} \] \[ = k q_1 \left(\frac{36 - 25}{900}\right) \hat{i} \] \[ = k q_1 \left(\frac{11}{900}\right) \hat{i} \] --- ### **Step 7: Draw the Force on \(q_3\)** - \(q_3 < 0\) is at \(P\). - The force on \(q_3\) is \(\vec{F}_3 = q_3 \vec{E}_{\text{total}}\). - Since \(\vec{E}_{\text{total}}\) is to the **right** (\(+x\)), and \(q_3\) is negative, the force will be to the **left** (\(-x\)). --- ### **Step 8: Final Answers** #### **Electric Field at \(P\):** \[ \boxed{ \vec{E}_P = k q_1 \frac{11}{900} \hat{i} } \] (Direction: \(+\hat{i}\), or to the right) #### **Force on \(q_3\):** \[ \vec{F}_3 = q_3 \vec{E}_P \] (Direction: \(-x\), to the left, since \(q_3 < 0\)) --- ### **Summary Table** | Quantity | Value/Direction | |----------------------------------|--------------------------------------| | \(\vec{E}_P\) | \(k q_1 \dfrac{11}{900} \hat{i}\) | | Direction of \(\vec{E}_P\) | \(+\hat{i}\) (to the right) | | Direction of force on \(q_3\) | \(-\hat{i}\) (to the left) | #### **On your diagram:** - Draw a rightward arrow at \(P\) for \(\vec{E}_P\). - Draw a leftward arrow at \(q_3\) for the force on \(q_3\). If you have the value for \(q_1\), you can substitute it in for a numerical answer.