We observe that the ball hangs at an angle of θ=18.0o from the vertical. what is the magnitude of the electric field.

Answer:

To answer this question, we need to use some concepts from electrostatics and trigonometry. First, we need to find the charge on the ball, which we can do by using the formula for the electric force between two point charges:

F=kr2q1q2

where k is the Coulomb’s constant, q1 and q2 are the charges, and r is the distance between them. In this case, we have:

F=kL2qQ

where q is the charge on the ball, Q is the charge on the rod, and L is the length of the string. We also know that the electric force is balanced by the tension in the string, which has a horizontal component equal to F and a vertical component equal to mg, where m is the mass of the ball and g is the acceleration due to gravity. Using trigonometry, we can write:

F=Tsinθ

mg=Tcosθ

where θ is the angle that the string makes with the vertical. Solving for T, we get:

T=cosθmg

Substituting this into the first equation, we get:

F=cosθmgsinθ

Simplifying, we get:

F=mgtanθ

Now we can solve for q, using the values given in the question:

q=kQFL2=kQmgL2tanθ

q=(9.0×109)(5.0×10−6)(0.050)(9.8)(0.50)2tan18.0∘

q=1.3×10−6C

Next, we need to find the electric field at the location of the ball, which we can do by using Gauss’s law for a sphere of charge1. Gauss’s law states that:

∮E⋅dA=ϵ0Qenc

where E is the electric field, dA is an infinitesimal area element on a closed surface, Qenc is the charge enclosed by the surface, and ϵ0 is the permittivity of free space. In this case, we can choose a spherical surface with radius r centered at the rod, which encloses both the rod and the ball. The electric field on this surface is radial and constant, so we can write:

E(4πr2)=ϵ0Q+q

where Q is the charge on the rod and q is the charge on the ball. Solving for E, we get:

E=4πϵ0r2Q+q

Now we can plug in the values given in the question:

E=4π(8.85×10−12)(0.50)2(5.0×10−6)+(1.3×10−6)

E=1.7×105N/C

Therefore, the magnitude of the electric field at the location of the ball is 170 kN/C.