Voltage vs Distance graph for a given E vs D graph

[pgirl1729]

Homework Statement

How do you draw the voltage vs distance graphs for the following Electric field intensity vs distance graphs?

Relevant Equations

V= (-)Ed

I just started learning field intensity and voltage. So when a voltage vs distance graph is given, I used to think of it as a displacement time graph and drew the respective voltage vs distance graph, as if I was drawing the relevant velocity time graph, but on the opposite sides. It worked for almost all the graph questions i did for V vs D up to now. But when I followed the same method for an E vs D graph shown below,

I have a graph that starts at 0,0 where y increases negatively as x increases positively, but the gradient continues to rise. But the answer given by my teacher, laid in the positive side, had a positive voltage initially and that kept decreasing up to a certain point, as x increased positively. the slope increased with distance.

Similar contradictions were found in the two graphs below as well.

Would prefer an answer without calculus.

 

[nasu]

Would prefer an answer without calculus.

What is the question?

 

[pgirl1729]

What is the question?

Q. Draw the Voltage vs distance graphs for the following Electric field intensity vs Distance graphs
1.

2.

3.

(sorry for the confusion)

 

[haruspex]

I have a graph that starts at 0,0 where y increases negatively as x increases positively, but the gradient continues to rise. But the answer given by my teacher, lied in the positive side, had a positive voltage initially and that kept decreasing up to a certain point, as x increased positively. the slope increased with distance.

I am not sure I understand either of those two descriptions. As I read them, you are saying that both have negative gradients increasing monotonically in magnitude, the only difference being that yours starts at zero and the given answer starts positive. If so, both are valid. What the starting voltage is depends on your reference.

 

[nasu]

You cannot treat this type of problem without concepts from calculus. You can do it without using the word "calculus". But the concepts of gradient, slope as derivative, etc are all calculus concepts, even if you don't call them so.
So, let's assume that you know/undertsand that the electric field is the negative of the gradient of the potential. For the 1D case this means that the electric field is equal the slope of the graph of the potential, with a minus sign.
The first graph then shows a positive, increasing electric field versus distance. So the potential should be a funtion with an increasing slope, but a negative slope. So a function decreasing faster and faster. It is actually an upside down parabola but you cannot find the exact shape without some notions of integrals.
Are you supposed to actually find the curve or just to pick the right one from a series of choices?

 

[pgirl1729]

You cannot treat this type of problem without concepts from calculus. You can do it without using the word "calculus". But the concepts of gradient, slope as derivative, etc are all calculus concepts, even if you don't call them so.
So, let's assume that you know/undertsand that the electric field is the negative of the gradient of the potential. For the 1D case this means that the electric field is equal the slope of the graph of the potential, with a minus sign.
The first graph then shows a positive, increasing electric field versus distance. So the potential should be a funtion with an increasing slope, but a negative slope. So a function decreasing faster and faster. It is actually an upside down parabola but you cannot find the exact shape without some notions of integrals.
Are you supposed to actually find the curve or just to pick the right one from a series of choices?

no choices. is it something like this?

with a value for V

 

[haruspex]

when a voltage vs distance graph is given, I used to think of it as a displacement time graph and drew the respective voltage vs distance graph, as if I was drawing the relevant velocity time graph, but on the opposite sides.

That analogy works, but it is not clear how you used it in this question. I assume you reversed it, so taking the given E/x graph as analogous to -v/t you drew the s/t graph corresponding to the latter, i.e. its integral.

no choices. is it something like this? View attachment 368163 with a value for V

Something like that, but you have not followed your analogy very accurately. A velocity that starts at zero then decreases linearly with time describes a dropped stone. What does the distance/time graph look like for that?

 

[pgirl1729]

Something like that, but you have not followed your analogy very accurately. A velocity that starts at zero then decreases linearly with time describes a dropped stone. What does the distance/time graph look like for that?

In a distance-time graph,

In a displacement-time graph same shape as

, but starts at 0,0 and continues on the negative side, if we assume displacement to be 0 initially.

 

[haruspex]

In a distance-time graph, View attachment 368164

In a displacement-time graph same shape as View attachment 368165 , but starts at 0,0 and continues on the negative side, if we assume displacement to be 0 initially.

To my eye, those two curves look different. The second is close to being a quadrant of a circle but the first exhibits changing curvature along its length.
What is the displacement / time equation for a dropped stone?

 

[nasu]

no choices. is it something like this? View attachment 368163 with a value for V

Yeah, qualitatively it may be, for the first graph of E is x. But these points (a and b) are from some other problem, right?

Also, as you know (I hope) the potential is not uniquely defined. So for a given E you can have a family of potential curves, shifted along the vertical axis.