The drawing shows a small sphere of mass m = 0.24 g that carries a charge of q = 8.5 × 10−10 C.

Question

The drawing shows a small sphere of mass m = 0.24 g that carries a charge of q = 8.5 × 10⁻¹⁰ C. The sphere is attached to one end of a very thin silk string of length L = 5.5 cm. The other end of the string is attached to a large vertical conducting plate that has a charge density of σ = 30 × 10⁻⁶ C/m².

I am not able to include the image for some reason but let me describe it to you:

there is a straight, uniform rod of + charges, then the silk string radiates out from the rod at and angle θ with the sphere attached to the bottom of the rod not attached to the rod.

1. Write an expression for the magnitude of the electric field due to the charge distribution on the conductive plate. 
2. Write an expression for the horizontal component of the net force acting on the charged sphere in terms of the tension T in the string, the electric field, and the given variables. 
3. Write an expression for the vertical component of the net force acting on the charged sphere in terms of the tension T in the string and the given variables. 
4. What is the angle in degrees that the string makes with respect to the vertical? 

Answer 1

The magnitude of the electric field for the vertical plane is E= σ/2ε₀ where ε₀ is the electric permittivity. The horizontal component of the net force is Eq - Tsinθ=0 where F=Eq and Tsinθ is the horizontal component of the tension in the string. The vertical force is the weight which is mg and this is offset by the vertical of the tension which is Tcosθ. So the equation is mg-Tcosθ =0. To find the angle θ divide the first equation by the second one to get tanθ =σq/2ε₀mg. Using the values given in the problem gives an angle around 31.47º.