Answer:
a) (5.59 × 10³) N/C
b) 0.226 N directed away from the spider.
Explanation:
a) Electric field, E, felt as a result of point charge, Q, at a distance, d away is given by
E = kQ/d²
So, magnitude of the electric field due to the charge on the second spider at the position of the first spider
Q = 4.2 µC = 4.2 × 10⁻⁶ C
k = Coulomb's constant = 8.99 × 10⁹ Nm²/C
d = 2.6 m
E = (8.99 × 10⁹ × 4.2 × 10⁻⁶)/2.6²
E = 5.59 × 10³ N/C
b) Tension in the silk fiber above the spider is the net force due to the weight of spider one and the force of repulsion due the two charges.
Force due to the two charges = Eq
where q now represents the charge of the first spider at the first point, feeling the electric field calculated in (a)
F = 5.59 × 10³ × 3.4 × 10⁻⁶ = 0.01901 N directed upwards. (That is, F = + 0.019 N)
Weight of the spider = mg = 0.025 × 9.8 = 0.245 N directed downwards. (That is, W = -0.245 N)
Net force, T = mg - F = 0.245 - 0.019 = 0.226 N (that is, 0.226 N, directed upwards, away from the spider).
