A spider begins to spin a web by first hanging from a ceiling by his fine, silk fiber. He has a mass of 0.025 kg and a charge of 3.4 µC. A second spider with a charge of 4.2 µC rests in her own web exactly 2.0 m vertically below the first spider. (a) What is the magnitude of the electric field due to the charge on the second spider at the position of the first spider? ___N/C (b) What is the tension in the silk fiber above the first spider? ___N

A) 9450 N

B) 0.213 N

Explanation:

A)

The magnitude of the electric field produced by a single-point charge is given by:

where:

is the Coulomb's constant

is the charge responsible for the field

r is the distance at which the field is calculated

In this problem:

is the  charge on the 2nd spider, producing the electric field and causing the force on the 1st spider

r = 2.0 m is the distance between the two spiders

Therefore, the electric field at the location of the 1st spider is:

2)

We observe that there are 3 forces acting on the 1st spider:

- The electric force , in the upward direction (both spiders are positively charge, so the force is repulsive, and since the 1st spider is above, the force on it is upward)

- The weight of the spider, W, in the downward direction

- The tension in the silk fiber, T, upward

Since the spider is in equilibrium we can write:

And writing explicitely each force:

where:

is the charge of the 1st spider

is the magnitude of the electric field

m = 0.025 kg is the mass of the spired

is the acceleration due to gravity

Solving for T, we find:

answer is slate

hope this