Answer:
a) F₂₁ = 0.02 N, attracting.
b) F₁₂ = 0.02 N, attracting.
Explanation:
a)
- The magnitude of the force that q₁ exerts on q₂ (F₂₁) is given by Coulomb's Law, as follows:
[tex]F_{21} = k * \frac{q_{1} *q_{2}}{r_{12}^{2} } = 9e9 N.m2/C2 * \frac{(25e-9C)*(75e-9C)}{(0.03m)^{2}} = 0.02 N (1)[/tex]
- Since q₁ and q₂ have opposite signs, the force between them will be always attractive, i.e., from q₂ towards q₁, along the line that joins both charges.
b)
- The magnitude of the force on q₁ due to q₂ can be obtained applying Newton's 3rd Law, or using (1), because all parameters are the same, so F₁₂ (in magnitude) = F₂₁ = 0.02 N
- As we have already said, it must be opposite to the one found in a) so it must go from q₁ towards q₂, it is an attracting force also.
