How Do You Calculate the Electric Charge of a Sphere?

[Ryo124]

Homework Statement

You have an electrically charged sphere.

The strength of an electric field 0.1 m away from the center of the sphere is 30,000 N/C.

The radius of the sphere is 0.05 m.

What is the electric charge (Q) of the sphere?

Homework Equations

E = F/q - E = -V/d

The Attempt at a Solution

I don't know how to attempt this problem, I'm not even sure if these equations are relevant. Someone please help.

 

[Ryo124]

Nevermind, I got it. 3.3E-8 C

 

[ogb p]

I can provide some guidance on how to approach this problem. First, it is important to understand that electric charge is a fundamental property of matter and is measured in coulombs (C). In order to solve for the electric charge of the sphere, we need to use the equation E = F/q, where E is the electric field strength, F is the electric force, and q is the electric charge. In this case, we are given the electric field strength (E) and the distance (0.1 m) from the center of the sphere, but we do not have the electric force (F). To solve for F, we can use the equation F = kQq/r^2, where k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2), Q is the electric charge of the sphere, q is the electric charge of the test charge, and r is the distance between the two charges. Since we are given the radius of the sphere (0.05 m), we can substitute this value for r in the equation.

Therefore, the equation becomes:

E = kQ/r^2

Solving for Q, we get:

Q = (Er^2)/k

Substituting the given values, we get:

Q = (30000 N/C x (0.05 m)^2) / (8.99 x 10^9 Nm^2/C^2)

Q = 0.0833 x 10^-9 C

Q = 8.33 x 10^-11 C

Therefore, the electric charge of the sphere is 8.33 x 10^-11 C. It is important to note that the negative sign in the equation E = -V/d is not relevant in this problem as we are only concerned with the magnitude of the electric field strength. I hope this helps in solving the problem.