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Electric field from spherically symmetric charge distributio

  • #1

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?

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
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
 

Answers and Replies

  • #2
follow up question:
Is there a hint buried somewhere in the problem that should tell me that the electric field should go as E=σ/2ε0, σ being charge/area?
If so where?
if not should I assume the electric field goes as E=KQ/r2?
 

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