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In the diagram below calculate the magnitude of that force that (a) q1 exerts on q2 (b) q2 exerts on q1.

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Part 10· 4 steps

In the diagram below calculate the magnitude of that force that (a) q1q_1 exerts on q2q_2 (b) q2q_2 exerts on q1q_1.

Given values from the diagram:
q1=−4.0μC=−4.0×10−6q_1 = -4.0 \mu C = -4.0 \times 10^{-6} C
q2=3.00μC=3.00×10−6q_2 = 3.00 \mu C = 3.00 \times 10^{-6} C
Horizontal distance, x=0.3x = 0.3 m
Vertical distance, y=0.4y = 0.4 m
Coulomb's constant, k=8.987×109 N m2/C2k = 8.987 \times 10^9 \text{ N m}^2/\text{C}^2

  1. Calculate the distance rr between q1q_1 and q2q_2 using the Pythagorean theorem.

    r=x2+y2r = \sqrt{x^2 + y^2} r=(0.3 m)2+(0.4 m)2r = \sqrt{(0.3 \text{ m})^2 + (0.4 \text{ m})^2} r=0.09 m2+0.16 m2r = \sqrt{0.09 \text{ m}^2 + 0.16 \text{ m}^2} r=0.25 m2r = \sqrt{0.25 \text{ m}^2} r=0.5 mr = 0.5 \text{ m}

  2. Use Coulomb's Law to calculate the magnitude of the electrostatic force between q1q_1 and q2q_2.

    F=k∣q1q2∣r2F = k \frac{|q_1 q_2|}{r^2}

  3. Substitute the values into the formula.

    F=(8.987×109 N m2/C2)∣(−4.0×10−6 C)(3.00×10−6 C)∣(0.5 m)2F = (8.987 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{|(-4.0 \times 10^{-6} \text{ C})(3.00 \times 10^{-6} \text{ C})|}{(0.5 \text{ m})^2}

  4. Calculate the force.

    F=(8.987×109)∣−1.2×10−11∣0.25F = (8.987 \times 10^9) \frac{|-1.2 \times 10^{-11}|}{0.25} F=(8.987×109)1.2×10−110.25F = (8.987 \times 10^9) \frac{1.2 \times 10^{-11}}{0.25} F=(8.987×109)×(4.8×10−11)F = (8.987 \times 10^9) \times (4.8 \times 10^{-11}) F=0.431376 NF = 0.431376 \text{ N}

    According to Newton's third law, the magnitude of the force q1q_1 exerts on q2q_2 is equal to the magnitude of the force q2q_2 exerts on q1q_1.

Part a

The magnitude of the force that q1q_1 exerts on q2q_2 is* 0.431 N\boxed{\mathbf{0.431 \text{ N}}}.

Part b

The magnitude of the force that q2q_2 exerts on q1q_1 is* 0.431 N\boxed{\mathbf{0.431 \text{ N}}}.

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