Electric flux and electric fields

[aiaigasa]

Hi, I'm having a bit of trouble with the following problem. Any insight as to what I'm doing wrong?

Homework Statement

A solid metal sphere with radius 0.900 m carries a net charge of 0.130 nC. Find the magnitude of the electric field at at a point inside the sphere, 0.100 m below the surface.

Homework Equations

$$\Phi = \int\vec{E}\cdot\vec{A} = \frac{Q}{\epsilon_{0}}$$

The Attempt at a Solution

$$Q = \frac{0.130 \times 10^{-9}}{\frac{4}{3}\pi(0.9)^{3}} \times \frac{4}{3}\pi(0.8)^{3} = 9.13032 \times 10^{-11}$$
$$\Phi = \frac{9.13032 \times 10^{-11}}{\epsilon_{0}} = 10.3119$$
$$E = \frac{\Phi}{A} = \frac{10.3119}{4\pi(0.8)^{2}} = 1.28217 \frac{N}{C}$$

 

[Defennder]

What can you deduce about a charged conductor under electrostatic equilibrium?

 

[Moetasim]

Hello, it seems like you are trying to solve the problem using the electric flux equation. However, this equation is used to calculate the total electric flux through a closed surface, not the electric field at a specific point. To find the electric field at a point inside the sphere, you can use the equation:

E = \frac{kQ}{r^2}

Where k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2), Q is the net charge of the sphere, and r is the distance from the center of the sphere to the point where you want to find the electric field. In this case, r = 0.1m. Plugging in the values, we get:

E = \frac{(8.99 \times 10^9)(0.130 \times 10^{-9})}{(0.1)^2} = 9.167 \times 10^3 \frac{N}{C}

This is the magnitude of the electric field at a point inside the sphere, 0.1m below the surface. I hope this helps!