# Electric field from spherically symmetric charge distributio

## Homework Statement

A spherically symmetric charge distribution produces the electric field E=(200/r)r(hat)N/C, where r is in meters.
a) what is the electric field strength at 10cm?
b)what is the electric flux through a 20cm diameter spherical surface that is concentric with the charge distribution?
c)How much charge is inside this 20cm diameter spherical surface?

## Homework Equations

φe=∫E⋅dA=Qenclosed0

## The Attempt at a Solution

question a is pretty straightforward:
a) E=(200/.1m)N/C=2x103N/C
part b is where I get stuck, heres my attempt:
∫E⋅dA=Qenclosed0, Qenclosed is just some charge distribution so lets say we know it and keep it in its variable form, Qenclosed.
∫E⋅dA=Qenclosed0→EA=Qenclosed0
since the 20cm diameter spherical surface we drew is a known shape (spherical) we can use it as a gaussian surface?
E4πr2=Qenclosed0→E=KQenclosed/r2. Clearly I see the problem of having two unknowns here.

I guess the problem says nothing about what the dimensions of the "spherical symmetric charge distribution" is... maybe this comes into play?

The problem states what the electric field goes as
so for part b:
E=(200/.1m)N/C, our r for part b is the radius of the 20cm concentric spherical surface is 10cm→.1m
E=2000N/C
Φe=EA=2000N/C(4π0.1m2)=251.3Nm2/C

If this is correct then part c should be easy:
φe=∫E⋅dA=EA=Qenclosed0
we just solved for fluxφe
φeε0=Qenclosed therefore,
Qenclosed=(251.3Nm2/C)(8.85x10^-12C2/Nm2)=2.24x10^-9C