Finding charge of insulating sphere and conductor

[phanman]

Homework Statement

For the configuration shown in the figure below, suppose that a = 6.10 cm, b = 21.6 cm, and c = 26.6 cm. Furthermore, suppose that the electric field at a point 11.5 cm from the center is 3.37x10^3 N/C radially inward, while the electric field at a point 58.6 cm from the center is 1.30x10^2 N/C radially outward.I can't copy picture so i will describe it. In the middle there is a sphere with radius a, it is an insulator. There is an outer ring around this sphere which has radius b. Than the last ring is the outermost ring with radius c. The last ring is a conductor.

(a) find the charge of the insulating sphere
(b) the net charge on the hollow conducting sphere.

Homework Equations

k_e(Coloumb constant) = 8.99 x 10^9 Nm^2/C^2
k_e(q/r^2)

The Attempt at a Solution

What i did was since the second ring there is an electric field of 3.37x10^3N/C which is given, i calculated q using the equation about. Than the innermost sphere charge should be the opposite of that.

 

[kuruman]

Hi phanman, welcome to PF.

There is no configuration "shown below." Without it, we cannot help you.

 

[tomcue]

I would approach this problem by first understanding the setup and what is being asked. From the given information, it seems that there is a configuration of three concentric rings - an insulating sphere in the middle, an outer ring, and an outermost ring which is a conductor. The electric field at two points, 11.5 cm and 58.6 cm from the center, is also given.

To find the charge of the insulating sphere (a), we can use the equation for electric field, E = k_e(q/r^2). We know the electric field at a point 11.5 cm from the center is 3.37x10^3 N/C radially inward, so we can set up the equation as: 3.37x10^3 = (8.99x10^9)(q/11.5^2). Solving for q, we get the charge of the insulating sphere to be -2.37x10^-7 C (since the electric field is radially inward, the charge must be negative).

To find the net charge on the hollow conducting sphere (b), we can use the same equation, E = k_e(q/r^2), but using the electric field at a point 58.6 cm from the center, which is 1.30x10^2 N/C radially outward. Setting up the equation, we get: 1.30x10^2 = (8.99x10^9)(q/58.6^2). Solving for q, we get the net charge on the hollow conducting sphere to be 3.69x10^-6 C.

It's important to note that in this problem, we assume the outermost ring is a conductor, so the electric field is constant throughout the surface of the ring. If it was not specified, we would need more information to calculate the charge on the outermost ring. Overall, it's important to carefully read and understand the problem and use the correct equations and assumptions to find the solution.