Regents Physics Worksheet 4.1.3 – Coulomb’s Law (20 points) – Show all work – multiple choice answers MUST be proven One object with a charge of +2.0 x 10^(-6) coulombs and another object with a charge of -3.0 x 10^(-6) coulombs are placed so that they are 10 centimeters apart. Determine the electrostatic force that the two objects exert on one another. In the diagram, the charge is represented as q1 and q2. Using Coulomb’s Law, the formula for the electrostatic force (F) between two objects is: F = k * (|q1| * |q2|) / r^2 where k is the electrostatic constant (k = 9 x 10^9 Nm^2/C^2), |q1| and |q2| are the magnitudes of the charges, and r is the distance between the objects. Plugging in the values: F = (9 x 10^9 Nm^2/C^2) * ((2.0 x 10^(-6) C) * (3.0 x 10^(-6) C)) / (0.1 m)^2 F = (9 x 10^9 Nm^2/C^2) * (6.0 x 10^(-12) C^2) / 0.01 m^2 F = (9 x 10^9 Nm^2/C^2) * 6.0 x 10^(-10) C^2 / 0.01 m^2 F = (9 x 10^9 Nm^2/C^2) * 6.0 x 10^(-8) N F = 54 x 10^1 N F = 540 N Therefore, the electrostatic force that the two objects exert on one another is 540 N. If the distance between the objects were doubled, what would the magnitude of the electrostatic force be? To find the new magnitude of the electrostatic force, we can use the inverse square relationship between force and distance. If the distance is doubled, the force will be reduced by a factor of 4. Therefore, the new magnitude of the electrostatic force would be 540 N / 4 = 135 N. Sketch a graph that relates the electrostatic force between two objects and the distance between them
The Correct Answer and Explanation is:
Correct Calculation and Explanation for Coulomb’s Law Problem
Given Data:
- Charge 1: q1=+2.0×10−6 Cq_1 = +2.0 \times 10^{-6} \, Cq1=+2.0×10−6C
- Charge 2: q2=−3.0×10−6 Cq_2 = -3.0 \times 10^{-6} \, Cq2=−3.0×10−6C
- Distance between objects: r=10 cm=0.1 mr = 10 \, cm = 0.1 \, mr=10cm=0.1m
- Electrostatic constant: k=9×109 Nm2/C2k = 9 \times 10^9 \, Nm^2/C^2k=9×109Nm2/C2
Coulomb’s Law Formula:F=k×∣q1∣×∣q2∣r2F = k \times \frac{|q_1| \times |q_2|}{r^2}F=k×r2∣q1∣×∣q2∣
Step by Step Solution:
Substitute the values into the formula:F=(9×109)×(2.0×10−6)×(3.0×10−6)(0.1)2F = (9 \times 10^9) \times \frac{(2.0 \times 10^{-6}) \times (3.0 \times 10^{-6})}{(0.1)^2}F=(9×109)×(0.1)2(2.0×10−6)×(3.0×10−6)
First, multiply the charges:2.0×10−6×3.0×10−6=6.0×10−122.0 \times 10^{-6} \times 3.0 \times 10^{-6} = 6.0 \times 10^{-12}2.0×10−6×3.0×10−6=6.0×10−12
Square the distance:(0.1)2=0.01 m2(0.1)^2 = 0.01 \, m^2(0.1)2=0.01m2
Now calculate the force:F=(9×109)×6.0×10−120.01=(9×109)×(6.0×10−10)=54 NF = (9 \times 10^9) \times \frac{6.0 \times 10^{-12}}{0.01} = (9 \times 10^9) \times (6.0 \times 10^{-10}) = 54 \, NF=(9×109)×0.016.0×10−12=(9×109)×(6.0×10−10)=54N
Final Correct Force:
The correct electrostatic force is 54 N, not 540 N as previously stated.
Effect of Doubling the Distance:
The force follows an inverse square law. If the distance doubles:rnew=2×0.1=0.2 mr_{\text{new}} = 2 \times 0.1 = 0.2 \, mrnew=2×0.1=0.2m
New force calculation:Fnew=k×∣q1∣×∣q2∣rnew2=(9×109)×6.0×10−12(0.2)2=(9×109)×6.0×10−120.04F_{\text{new}} = k \times \frac{|q_1| \times |q_2|}{r_{\text{new}}^2} = (9 \times 10^9) \times \frac{6.0 \times 10^{-12}}{(0.2)^2} = (9 \times 10^9) \times \frac{6.0 \times 10^{-12}}{0.04}Fnew=k×rnew2∣q1∣×∣q2∣=(9×109)×(0.2)26.0×10−12=(9×109)×0.046.0×10−12
Simplifying:Fnew=(9×109)×1.5×10−10=13.5 NF_{\text{new}} = (9 \times 10^9) \times 1.5 \times 10^{-10} = 13.5 \, NFnew=(9×109)×1.5×10−10=13.5N
New Force:
The new force is 13.5 N when the distance doubles.
Graph Description:
The relationship between electrostatic force and distance is an inverse square curve. As distance increases, force decreases rapidly. At small distances, force is large; as distance increases, force approaches zero but never becomes negative.
Graph Axes:
- X-axis: Distance rrr (meters)
- Y-axis: Electrostatic Force FFF (Newtons)
The curve starts high near the Y-axis and decreases sharply as rrr increases, forming a hyperbolic shape.
