What is the relationship between electric potential and electric field?

[mathguy831]

Q1: Use the graph (see attachment) to write the empirical equation V(r) for a charge cylindrical shell with appropriate constants, including units. Show all work.

Q2: What is the interpretation of the horizontal axis intercept?

Q3: What is the significance of the slope of the graph?

I know I have to start Q1 using Gauss's Law, but what to do from that there is beyond me. The cylindrical shell that I obtained this information from what negatively charged on the inner electrode and positively charged on the outer electrode. Thus as I measured the potentials moving farther from the negatively charged electrode, outward, to the positively charged electrode, I was able to record and graph the data you see in the attachment. Thank you in advance.

 

[Simon Bridge]

I know I have to start Q1 using Gauss's Law, but what to do from that there is beyond me.

... that is because you don't have to use Gauss' Law.
You are asked to find the equation of V vs r. You are given a graph of r vs V.

Note: r, in this case, is not a radius.

 

[mathguy831]

Okay, I graphed it wrong then I'll work on fixing that. But what about deriving an empirical equation. I'm still having trouble getting started on what to do.

 

[Simon Bridge]

When you plot your graph of V vs r, what type of curve do you get?
What is the equation for that sort of curve?

 

[mathguy831]

So when plotting the graph correctly this time it still formed a straight line, but with a steeper slope. So from there would I find the slope using (y2 - y1)/(x2 - x1) and then work that into the equation for a line y = mx + b?

(10 - 2)/(0.06 - 0.03) = 266.7 V/m

Then solving for b in the equation y = mx +b

y = b when x = 0

So looking at the point V = 2 and r = 0.03, when V drops by 2V, r drops by 2*0.00375 = 0.0075 m.

When V(r) = 0 , r = r = 0.03 - 0.0075 = 0.0225m

Therefore b = 0.0225m

V = 266.7 V(r) * r + 0.0225
So the empirical equation would be:
V(r) = (V - 0.0225m)/(266.7 * r)

?? I feel like I went wrong somewhere as this equation doesn't quite make sense to me.

Also what would the horizontal axis then represent? and would the slope of the line on the graph represent the magnitude of the E-field?

 

[Simon Bridge]

If you have (0,0) on your graph, you can just draw the best-fit line back and read the y-intercept value off the axis.

V = 266.7 V(r) * r + 0.0225
So the empirical equation would be:
V(r) = (V - 0.0225m)/(266.7 * r)

... neither of these is the equation of a straight line.
You just said that V vs r is a straight line! Thus: V(r)=mr+c

Also what would the horizontal axis then represent?

What does the horizontal axis normally represent for a graph?
In a graph of y vs x, the horizontal axis is the x-axis - it represents the condition that y=0.
In this case, the vertical axis has a physical meaning, it's the electric potential difference.
Difference from what?

... and would the slope of the line on the graph represent the magnitude of the E-field?

... you should have some notes on the relationship between the electric potential and the electric field that can tell you that.