A) 9450 N
B) 0.213 N
Explanation:
A)
The magnitude of the electric field produced by a single-point charge is given by:
[tex]E=k\frac{Q}{r^2}[/tex]
where:
[tex]k=8.99\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant
[tex]Q[/tex] is the charge responsible for the field
r is the distance at which the field is calculated
In this problem:
[tex]Q=4.2\mu C=4.2\cdot 10^{-6} C[/tex] 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:
[tex]E=(8.99\cdot 10^{9})\frac{4.2\cdot 10^{-6}C}{2.0^2}=9450 N[/tex]
2)
We observe that there are 3 forces acting on the 1st spider:
- The electric force [tex]F_E[/tex], 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:
[tex]F_E+T-W=0[/tex]
And writing explicitely each force:
[tex]qE+T-mg=0[/tex]
where:
[tex]q=3.4 \mu C = 3.4\cdot 10^{-6}C[/tex] is the charge of the 1st spider
[tex]E=9450 N[/tex] is the magnitude of the electric field
m = 0.025 kg is the mass of the spired
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
Solving for T, we find:
[tex]T=mg-qE=(0.025)(9.8)-(3.4\cdot 10^{-6})(9450)=0.213 N[/tex]
