Consider the arrangement of point charges in the figure.

Consider the arrangement of point charges in the figure. qc qd Othcexpertta Part (a) Determine the direction of the force on q (with q > 0) in the figure, given that qa = 7.50 ÂµC and qc = -7.50 ÂµC. Straight downward? Correct? Part (b) Calculate the magnitude of the force on the charge q in newtons, given that the square is 5 cm on each side and q = 2.25 ÂµC. Grade Summary Deductions Polcnia

The Correct Answer and Explanation is:

Given:

qa=qb=+7.50 μC=+7.50×10−6 Cq_a = q_b = +7.50 \, \mu\text{C} = +7.50 \times 10^{-6} \, \text{C}qa=qb=+7.50μC=+7.50×10−6C
qc=qd=−7.50 μC=−7.50×10−6 Cq_c = q_d = -7.50 \, \mu\text{C} = -7.50 \times 10^{-6} \, \text{C}qc=qd=−7.50μC=−7.50×10−6C
q=+2.25 μC=+2.25×10−6 Cq = +2.25 \, \mu\text{C} = +2.25 \times 10^{-6} \, \text{C}q=+2.25μC=+2.25×10−6C
Side length of square = L=7.00 cm=0.070 mL = 7.00 \, \text{cm} = 0.070 \, \text{m}L=7.00cm=0.070m
k=8.99×109 N⋅m2/C2k = 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2k=8.99×109N⋅m2/C2

Step 1: Geometry and Setup

The charge qqq is at the center of a square. Each corner has a point charge. Since the distance from center to any corner is:r=L2=0.0702≈0.0495 mr = \frac{L}{\sqrt{2}} = \frac{0.070}{\sqrt{2}} \approx 0.0495 \, \text{m}r=2L=20.070≈0.0495m

Step 2: Coulomb’s Law

The force between two charges is:F=k∣q1q2∣r2F = k \frac{|q_1 q_2|}{r^2}F=kr2∣q1q2∣

Let’s calculate the force magnitude due to any one corner charge on qqq:F=(8.99×109)(7.50×10−6)(2.25×10−6)(0.0495)2≈6.18 NF = \frac{(8.99 \times 10^9) (7.50 \times 10^{-6}) (2.25 \times 10^{-6})}{(0.0495)^2} \approx 6.18 \, \text{N}F=(0.0495)2(8.99×109)(7.50×10−6)(2.25×10−6)≈6.18N

This is the magnitude of force from each corner charge on qqq.

Step 3: Vector Components and Net Force

Let’s consider the symmetry:

qaq_aqa and qbq_bqb are positive → they repel qqq
qcq_cqc and qdq_dqd are negative → they attract qqq

Each force acts along the line connecting the center to the corner, at 45° to the axes.

Use symmetry and components:

Horizontal components cancel.
Vertical components add up downward.

Each force contributes a vertical component of:Fy=F⋅sin⁡(45∘)=F⋅22F_y = F \cdot \sin(45^\circ) = F \cdot \frac{\sqrt{2}}{2}Fy=F⋅sin(45∘)=F⋅22

So net force:Fnet=4⋅Fy=4⋅(6.18⋅22)=4⋅(6.18⋅0.7071)≈17.48 NF_{\text{net}} = 4 \cdot F_y = 4 \cdot \left(6.18 \cdot \frac{\sqrt{2}}{2}\right) = 4 \cdot (6.18 \cdot 0.7071) \approx 17.48 \, \text{N}Fnet=4⋅Fy=4⋅(6.18⋅22)=4⋅(6.18⋅0.7071)≈17.48N

✅ Final Answer:

F=17.48 N\boxed{F = 17.48 \, \text{N}}F=17.48N

📘 Explanation (300 words):

This problem involves calculating the net electric force acting on a charge qqq located at the center of a square with four surrounding charges at the corners. This setup is a classic application of Coulomb’s Law and requires vector addition due to the two-dimensional arrangement.

We begin by noting the nature and placement of the charges: two of the corner charges are positive, and two are negative, arranged diagonally across from each other. Because qqq is a positive test charge at the center, it experiences repulsion from the positive corner charges and attraction to the negative corner charges. The symmetry of the square ensures that horizontal force components from opposite charges cancel each other out, while vertical components reinforce one another. Thus, the total force is directed straight downward, as verified in part (a).

To calculate the magnitude of the force from each corner, we use Coulomb’s Law:F=k∣q1q2∣r2F = k \frac{|q_1 q_2|}{r^2}F=kr2∣q1q2∣

The distance rrr from the center to a corner is derived from geometry as L/2L/\sqrt{2}L/2, where LLL is the side length of the square.

Each charge exerts the same magnitude of force on qqq, but the direction varies. When breaking these into components, only the vertical components (aligned along the same line) add constructively, resulting in the net force.

After computing the component forces and summing, the final result for the net force on qqq is approximately 17.48 newtons downward. This illustrates how vector resolution and geometric symmetry simplify otherwise complex multi-charge problems in electrostatics.