- #1

cookiemnstr510510

- 162

- 14

## 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

φ

_{e}=∫E⋅dA=Q

_{enclosed}/ε

_{0}

## The Attempt at a Solution

question a is pretty straightforward:

a) E=(200/.1m)N/C=2x10

^{3}N/C

part b is where I get stuck, here's my attempt:

∫E⋅dA=Q

_{enclosed}/ε

_{0}, Q

_{enclosed}is just some charge distribution so let's say we know it and keep it in its variable form, Q

_{enclosed}.

∫E⋅dA=Q

_{enclosed}/ε

_{0}→EA=Q

_{enclosed}/ε

_{0}

since the 20cm diameter spherical surface we drew is a known shape (spherical) we can use it as a gaussian surface?

E4πr

^{2}=Q

_{enclosed}/ε

_{0}→E=KQ

_{enclosed}/r

^{2}. 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?

Just had an idea:

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

Φ

_{e}=EA=2000N/C(4π0.1m

^{2})=251.3Nm

^{2}/C

If this is correct then part c should be easy:

φ

_{e}=∫E⋅dA=EA=Q

_{enclosed}/ε

_{0}

we just solved for fluxφ

_{e}

φ

_{e}ε

_{0}=Q

_{enclosed}therefore,

Q

_{enclosed}=(251.3Nm

^{2}/C)(8.85x10^-12C

^{2}/Nm

^{2})=2.24x10^-9C