In the figure shown, if a = 3.0 mm, b = 4.0 mm, Q1 = 60 nC, Q2 = 80 nC, and q = 24 nC, what is the magnitude of the electric force on q?
Answers
Answer:
If a=3.0mm, b=4.0 mm, Q1= 60 nC, Q2= 80 nC and q= 32nC in the figure below, what is the magnitude of the total electric force on q ?
The answer is 1.3, what is the full procedure?
magnitude of force exerted by Q1 on q, F1 = k*Q1*q/(a^2 +b^2)
= 9*10^9*60*10^-9*32*10^-9/(0.003^2 + 0.004^2)
= 0.6912 N
direction of F1, theta1 = tan^-1(a/b)
= tan^-1(3/4)
= 36.9 degrees below +x axis
magnitude of force exerted by Q2 on q, F2 = k*Q2*q/(a^2 +b^2)
= 9*10^9*80*10^-9*32*10^-9/(0.003^2 + 0.004^2)
= 0.9216 N
direction of F2, theta2 = tan^-1(a/b)
= tan^-1(3/4)
= 36.9 degrees above +x axis
so, the angle between F1 and F2, theta = theta1 + theta2
= 36.9 + 36.9
= 73.8 degrees
the magnitude of the total electric force on q,
|Fnet| = sqrt(F1^2 + F2^2 + 2*F1*F2*cos(theta))
= sqrt(0.6912^2 + 0.9216^2 + 2*0.6912*0.9216*cos(73.8))
= 1.30 N
Given: a = 3.0 mm, b = 4.0 mm, Q1 = 60 nC, Q2 = 80 nC, and q = 24 nC
To find: magnitude of the electric force on q
Solution:
Magnitude of force exerted by Q₁ on q,
F₁ = k x Q₁ x q/(a² +b²)
= 9 × 10⁹ x 60 x 10⁻⁹ x 32 x 10⁻⁹ / (0.003² + 0.004²)
= 0.6912 N
Direction of F₁,
Θ₁ = tan⁻¹ (a/b)
= tan⁻¹ (3/4)
= 36.9 degrees below +x axis
Magnitude of force exerted by Q₂ on q,
F₂ = k x Q₂ x q / (a² +b²)
= 9 x 10⁹ x 80 x 10⁻⁹ x 32 x 10⁻⁹ / (0.003² + 0.004²)
= 0.9216 N
Direction of F₂,
Θ₂ = tan⁻¹ (a/b)
= tan⁻¹ (3/4)
= 36.9 degrees above +x axis
So, the angle between F₁ and F₂,
Θ = Θ₁ + Θ₂
= 36.9 + 36.9
= 73.8 degrees
The magnitude of the total electric force on q,
| Fnet | = sqrt (F₁² + F₂² + 2 x F₁ x F₂ x cos(Θ))
= sqrt (0.6912² + 0.9216² + 2 x 0.6912 x 0.9216 x cos(73.8))
= 1.30 N
The magnitude of the electric force on q is 1.3 N.