Gauss' Law - Electric Flux and Electric Field

AI Thread Summary
The discussion revolves around calculating electric flux and electric field for a uniformly charged hollow sphere. The electric flux is determined using the formula, yielding approximately 2.26 N*m²/C. For the electric field at a distance of 200 cm, the calculation using the appropriate formula results in a value of 8.99 x 10^9 N/C. It is clarified that the radius of the sphere does not affect the electric field outside the sphere, as it behaves like a point charge. Participants seek verification of these calculations and formulas used.
Yuxtapuesto
Messages
3
Reaction score
0

Homework Statement



A 2 C charge is spread uniformly in a hollow sphere with a radius of 50 cm.

a) What is the electric flux?

b) What is the magnitude and direction of an electric field 200 cm away?

Homework Equations



Electric flux = q/(permittivity of free space)

E = q/4(pi)(permittivity of free space)r2

q = charge
The permittivity of free space = 8.85 x 10-12 C2/(N*m2)
r = radius

The Attempt at a Solution



Because the electric flux = the enclosed net charge divided by the permittivity of free space (approx. 8.85 x 10-12) I attempted to use this equation. Using 2 C as the net charge, I found the answer to be about 2.26 N*m2/C

I had no idea how to solve the second part of the question, but referenced my book and found this equation:

E = q/4(pi)(permittivity of free space)r2

q = charge
The permittivity of free space = 8.85 x 10-12 C2/(N*m2)
r = radius

My problem with this equation is that it doesn't seem to incorporate both the 50 cm and the 200 cm distances that are given within the question, but using 200 cm as the value of r, I calculated the answer to part b as 8.99 x 109 N/C

Any help would be greatly appreciated!
 
Last edited:
Physics news on Phys.org
The reason it doesn't incorporate the 50cm i.e. the radius of the sphere is that outside the sphere the electric field is the same as if you concentrated the charge spread throughout the sphere into a point charge at the center of the sphere.

This can be visualised easily by the first equation you gave. The one for the flux. (We are talking about a reference point outside of the sphere) You said that the flux is proportional to the enclosed net charge, hence if it was the flux of a point charge 2C you would get the same result, since the net charge enclosed is the same, as of the sphere.
 
Thank you. Can you / anyone possibly verify these answers?
 
What do you mean by verifying them? you theory is ok... You just have to substitute thenumerical values...
 
I mean that if someone could solve this problem independent of my work and share his or her answer with me, I would greatly appreciate it.
 
What is the formula for electric flux
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top