This worksheet breaks down Coulomb's Law calculations, showing how to determine the magnitude and direction of electric forces between multiple point charges.

Here is the step-by-step solution to the problems presented in the worksheet.

Part a: Force on $q_2$ due to $q_1$

Goal: Find the magnitude and direction of the electric force exerted on charge $q_2$ by charge $q_1$.

Given:
* $q_1 = +1.5 \, \mu\text{C} = +1.5 \times 10^{-6} \, \text{C}$
* $q_2 = -0.5 \, \mu\text{C} = -0.5 \times 10^{-6} \, \text{C}$
* Distance $d = 0.5 \, \text{m}$
* Coulomb's Constant $k \approx 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2$

Step 1: Determine Direction
Since $q_1$ is positive and $q_2$ is negative, they attract each other. The force on $q_2$ pulls it towards $q_1$. Looking at the diagram, $q_1$ is to the left of $q_2$, so the force points to the left (negative x-direction).

Step 2: Calculate Magnitude
We use Coulomb's Law formula for magnitude: $F = k \frac{|q_1 q_2|}{d^2}$

$$F = (8.99 \times 10^9) \frac{(1.5 \times 10^{-6})(0.5 \times 10^{-6})}{(0.5)^2}$$

$$F = (8.99 \times 10^9) \frac{0.75 \times 10^{-12}}{0.25}$$

$$F = (8.99 \times 10^9) (3.0 \times 10^{-12})$$

$$F = 26.97 \times 10^{-3} \, \text{N} = 0.02697 \, \text{N}$$

Rounding to 3 significant figures gives $0.0270 \, \text{N}$.

Conclusion for a:
The force is 0.0270 N pointing to the left.

***

Part b: Force on $q_3$ (placed to the right of $q_2$)

Goal: Find the net force on a new charge $q_3$.

Given:
* $q_3 = -1.0 \, \mu\text{C} = -1.0 \times 10^{-6} \, \text{C}$
* Position: $0.25 \, \text{m}$ to the right of $q_2$.
* Distance from $q_2$ to $q_3$ ($d_{23}$) = $0.25 \, \text{m}$.
* Distance from $q_1$ to $q_3$ ($d_{13}$) = $0.5 \, \text{m} + 0.25 \, \text{m} = 0.75 \, \text{m}$.

Step 1: Analyze Forces
* Force from $q_1$ on $q_3$: $q_1$ is (+) and $q_3$ is (-). They attract. The force pulls $q_3$ to the left.
* Force from $q_2$ on $q_3$: $q_2$ is (-) and $q_3$ is (-). They repel. The force pushes $q_3$ away from $q_2$, which is to the right.

Let's define "Right" as positive ($+$) and "Left" as negative ($-$).

Step 2: Calculate Individual Forces

* Force from $q_1$ ($F_{13}$):
$$F_{13} = - k \frac{|q_1 q_3|}{d_{13}^2} = -(8.99 \times 10^9) \frac{(1.5 \times 10^{-6})(1.0 \times 10^{-6})}{(0.75)^2}$$
$$F_{13} = -(8.99 \times 10^9) \frac{1.5 \times 10^{-12}}{0.5625} \approx -0.0240 \, \text{N}$$

* Force from $q_2$ ($F_{23}$):
$$F_{23} = + k \frac{|q_2 q_3|}{d_{23}^2} = +(8.99 \times 10^9) \frac{(0.5 \times 10^{-6})(1.0 \times 10^{-6})}{(0.25)^2}$$
$$F_{23} = +(8.99 \times 10^9) \frac{0.5 \times 10^{-12}}{0.0625} \approx +0.0719 \, \text{N}$$

Step 3: Sum the Forces
$$F_{net} = F_{13} + F_{23}$$
$$F_{net} = -0.0240 \, \text{N} + 0.0719 \, \text{N} = +0.0479 \, \text{N}$$

Conclusion for b:
The net force is 0.0479 N pointing to the right.

***

Part c: Force on $q_3$ (halfway between $q_1$ and $q_2$)

Goal: Find the net force on $q_3$ placed in the middle.

Given:
* $q_3$ is halfway. Total distance is $0.5 \, \text{m}$, so distance to each charge is $0.25 \, \text{m}$.

Step 1: Analyze Forces
* Force from $q_1$ on $q_3$: Attraction (+ and -). Pulls $q_3$ to the left (towards $q_1$).
* Force from $q_2$ on $q_3$: Repulsion (- and -). Pushes $q_3$ away from $q_2$, which is also to the left.

Since both forces point left, we add their magnitudes and assign a negative sign.

Step 2: Calculate
$$F_{net} = - \left( k \frac{|q_1 q_3|}{d^2} + k \frac{|q_2 q_3|}{d^2} \right)$$
Factor out common terms ($k, q_3, d^2$):
$$F_{net} = - \frac{k |q_3|}{d^2} (|q_1| + |q_2|)$$
$$F_{net} = - \frac{(8.99 \times 10^9)(1.0 \times 10^{-6})}{(0.25)^2} (1.5 \times 10^{-6} + 0.5 \times 10^{-6})$$
$$F_{net} = - \frac{8990}{0.0625} (2.0 \times 10^{-6})$$
$$F_{net} = - (143840) (2.0 \times 10^{-6})$$
$$F_{net} \approx -0.2877 \, \text{N}$$

Rounding to 3 significant figures gives $-0.288 \, \text{N}$.

Conclusion for c:
The net force is 0.288 N pointing to the left.

***

Part d: Force on $q_3$ (moved downward)

Goal: Find the vector force on $q_3$ when it forms an isosceles triangle with $q_1$ and $q_2$.

Given:
* Horizontal separation between $q_1$ and $q_2$ is $0.5 \, \text{m}$.
* $q_3$ is equidistant to both. This means $q_3$ is on the perpendicular bisector of the line connecting $q_1$ and $q_2$.
* The problem states $q_3$ is moved $0.3 \, \text{m}$ downward. Let's assume this refers to the vertical leg of the triangle formed by the center point and $q_3$.
* Horizontal distance from center to $q_1$ (or $q_2$) = $0.25 \, \text{m}$.
* Vertical distance to $q_3$ = $0.3 \, \text{m}$.
* Distance $r$ from $q_1$ (or $q_2$) to $q_3$: Using Pythagoras ($a^2+b^2=c^2$), $r = \sqrt{0.25^2 + 0.3^2} = \sqrt{0.0625 + 0.09} = \sqrt{0.1525} \approx 0.3905 \, \text{m}$.

*Note: The provided answer key in the image uses $d=0.30$m in the denominator for the y-component calculation, implying a simplification or a specific interpretation where the distance used for the vertical component projection is treated differently. However, strictly following vector decomposition:*

Let's look at the symmetry:
1. X-components:
* $q_1$ pulls $q_3$ left and down.
* $q_2$ pushes $q_3$ left and down.
* Wait, let's re-evaluate directions carefully.
* $q_1$ (+) attracts $q_3$ (-). Vector points from $q_3$ toward $q_1$ (Up and Left).
* $q_2$ (-) repels $q_3$ (-). Vector points away from $q_2$ (Down and Left).
* Actually, usually in these "equidistant" problems, if $q_3$ is on the bisector, the horizontal components often cancel or add up depending on signs.
* Let's check the Answer Key logic provided in the image text: It lists an X-component of -0.288 N. This is exactly the same number as Part C. This implies that for the X-component, the problem might be treating the distances effectively as the horizontal separation or there is a specific geometric setup intended where the horizontal forces sum to the value in Part C.
* However, physically, if $q_3$ is $0.3$m down, the distance $r$ is larger than $0.25$m, so the force should be weaker than Part C.
* Let's look at the Y-calculation in the image: It calculates a small downward force.

Let's stick to the standard physics derivation which matches the final numbers in the image key:

X-Direction:
Due to the specific geometry and charges, the horizontal components of the attraction from $q_1$ and repulsion from $q_2$ both point to the left.

Y-Direction:
* Force from $q_1$ (attraction) has an upward vertical component.
* Force from $q_2$ (repulsion) has a downward vertical component.
* Since $q_1$ ($1.5 \mu C$) is stronger than $q_2$ ($0.5 \mu C$), the upward pull from $q_1$ is stronger than the downward push from $q_2$.
* Therefore, the net Y-force should be Upward (Positive).

*Correction based on Image Key:* The image key shows a negative J component ($-0.010 \hat{j}$). Let's re-read the diagram.
If $q_3$ is below the axis:
- Vector $q_1 \to q_3$ (Attraction): Points Up-Left.
- Vector $q_2 \to q_3$ (Repulsion): Points Down-Left.

Let's calculate the Y-components explicitly using $r \approx 0.39$m.
$\sin(\theta) = \frac{0.3}{0.3905} \approx 0.768$
$\cos(\theta) = \frac{0.25}{0.3905} \approx 0.640$

$F_{1y} (\text{from } q_1) = + k \frac{q_1 q_3}{r^2} \sin(\theta)$ (Upward)
$F_{2y} (\text{from } q_2) = - k \frac{q_2 q_3}{r^2} \sin(\theta)$ (Downward)

$F_{net,y} = \frac{k q_3 \sin(\theta)}{r^2} (q_1 - q_2)$? No, magnitudes.
$F_{net,y} = \frac{8.99e9 \cdot 1e-6 \cdot 0.768}{0.1525} (1.5e-6 - 0.5e-6)$
$F_{net,y} = \frac{6904}{0.1525} (1.0e-6) \approx 0.045 \, \text{N}$ Upward.

*Discrepancy Note:* The provided solution in the image yields $-0.010 \hat{j}$. This result in the image key appears to rely on a different geometric interpretation or contains a calculation error regarding the angle/distance. Specifically, the image key calculates the Y-term using $(0.30)^2$ in the denominator directly, which is physically incorrect for Coulomb's law (which requires the square of the hypotenuse distance, not just the vertical leg).

However, as an educational assistant, I must provide the correct physics answer while acknowledging the context. The most robust answer follows the vector addition method.

Recalculating based on strict physics principles:

1. Distance $r$: $\sqrt{0.25^2 + 0.3^2} = 0.3905 \, \text{m}$.
2. Magnitude of Force from $q_1$:
$F_1 = k \frac{|q_1 q_3|}{r^2} = 8.99e9 \frac{1.5e-6 \cdot 1e-6}{0.1525} = 0.0882 \, \text{N}$.
3. Magnitude of Force from $q_2$:
$F_2 = k \frac{|q_2 q_3|}{r^2} = 8.99e9 \frac{0.5e-6 \cdot 1e-6}{0.1525} = 0.0294 \, \text{N}$.
4. Components:
Angle $\theta$ with horizontal: $\tan^{-1}(0.3/0.25) = 50.19^\circ$.
$\cos(50.19^\circ) = 0.640$, $\sin(50.19^\circ) = 0.768$.

X-Components (Both Left/Negative):
$F_{1x} = -F_1 \cos(\theta) = -0.0882 \cdot 0.640 = -0.0564 \, \text{N}$.
$F_{2x} = -F_2 \cos(\theta) = -0.0294 \cdot 0.640 = -0.0188 \, \text{N}$.
$F_{net,x} = -0.0564 - 0.0188 = \mathbf{-0.0752 \, \text{N}}$.

Y-Components:
$F_{1y}$ (Attraction to $q_1$ which is Up-Left) $= +F_1 \sin(\theta) = +0.0882 \cdot 0.768 = +0.0677 \, \text{N}$.
$F_{2y}$ (Repulsion from $q_2$ which is Down-Left) $= -F_2 \sin(\theta) = -0.0294 \cdot 0.768 = -0.0226 \, \text{N}$.
$F_{net,y} = 0.0677 - 0.0226 = \mathbf{+0.0451 \, \text{N}}$.

*Self-Correction*: The answers provided in the worksheet image (Part D) seem to contain significant errors in methodology (using vertical distance instead of hypotenuse for Coulomb's law denominator). As a tutor, I should provide the scientifically accurate result derived above, but note that it differs from the potentially flawed key in the image.

However, looking closely at the prompt's request "Solve the problem accurately", I will provide the standard correct physics solution.

Final Calculation Summary:
* X-component: Both charges push/pull $q_3$ to the left. Net X force is approx -0.075 N.
* Y-component: The stronger positive charge $q_1$ pulls up harder than the weaker negative charge $q_2$ pushes down. Net Y force is approx +0.045 N.

*(Note: If you are required to match the worksheet's specific erroneous method exactly, the answer would be $-0.288 \hat{i} - 0.010 \hat{j}$, but this is physically incorrect because it ignores the change in distance $r$ and vector angles properly.)*

Given the instruction to solve accurately, I will provide the physically correct values.

Final Answer:
a. Magnitude: 0.0270 N, Direction: Left (towards $q_1$)
b. Magnitude: 0.0479 N, Direction: Right (away from $q_2$)
c. Magnitude: 0.288 N, Direction: Left (towards $q_1$)
d. Magnitude: 0.088 N, Direction: Left and Up (Vector: $\approx -0.075 \hat{i} + 0.045 \hat{j}$ N)