Calculating Work and E-field from and equipotential contour graph

Click For Summary

Homework Help Overview

The discussion revolves around calculating work done by an external agent in moving a charge and determining the electric field magnitude using an equipotential contour graph. The subject area includes concepts of electric potential and electric fields.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss how to interpret equipotential lines to determine the potential difference (∆V) and the distance (∆L) for calculations. There is uncertainty about how to apply the equations for work and electric field based on the graph provided.

Discussion Status

Some participants have offered guidance on identifying potentials at specific points on the graph, while others question the implications of non-symmetrical contour lines on potential differences. Multiple interpretations of the graph and its implications for calculations are being explored.

Contextual Notes

There is mention of a specific potential difference of ±1 Volt between adjacent equipotential lines, which may influence the calculations. The original poster expresses confusion regarding the interpretation of the graph and the implications of contour line symmetry.

aximwolf
Messages
14
Reaction score
0

Homework Statement



prob04a_threeqcontour.gif


A) Calculate the work performed by an external agent to move a charge of -0.59x10-12 C from `i' to `b'.

B) Calculate the magnitude of the electric field at `k'

Homework Equations



W = q∆V

Where ∆V is the potential difference between the two potential lines each point
lies on

E = -∆V/∆L

The Attempt at a Solution



A) I know the work equation is charge x the change in V but I am not sure how to interpret the Equipotential lines in order to get change in V.

B) I know the equation is E = -∆V/∆L but I am not sure how to calculate both those variables from the graph.

The real question I have is how to interpret this graph to get ∆V and ∆L? I know that an Equipotential contour graph shows the plain perpendicular to the e-filed thus making Volts zero at any place there isn't a point charge.
 
Physics news on Phys.org
The difference in elctric potential for two adjacent equi-potential "lines" is ±1 Volt.
 
aximwolf said:

Homework Statement



prob04a_threeqcontour.gif


A) Calculate the work performed by an external agent to move a charge of -0.59x10-12 C from `i' to `b'.

B) Calculate the magnitude of the electric field at `k'

Homework Equations



W = q∆V

Where ∆V is the potential difference between the two potential lines each point
lies on

E = -∆V/∆L

The Attempt at a Solution



A) I know the work equation is charge x the change in V but I am not sure how to interpret the Equipotential lines in order to get change in V.

B) I know the equation is E = -∆V/∆L but I am not sure how to calculate both those variables from the graph.

The real question I have is how to interpret this graph to get ∆V and ∆L? I know that an Equipotential contour graph shows the plain perpendicular to the e-filed thus making Volts zero at any place there isn't a point charge.

(A)
You have the right idea, you just need to find the potentials at points i and b in order to calculation ΔV

For point i, notice that it is between the lines labelled 0V and +5. (Had you noticed the "+5" contour line?). How many lines over from the 0V line is i? How many volts is that?
 
Redbelly98 said:
(A)
You have the right idea, you just need to find the potentials at points i and b in order to calculation ΔV

For point i, notice that it is between the lines labelled 0V and +5. (Had you noticed the "+5" contour line?). How many lines over from the 0V line is i? How many volts is that?

will that work ... i mean the lines are not symmetrical ... if they were we could say that pot. increase by +1 when you move every line ahead ... can we do that in this case also when lines are not symmetrically separated?
 
Thank you tutors for your help that worked!
 
cupid.callin said:
will that work ... i mean the lines are not symmetrical ... if they were we could say that pot. increase by +1 when you move every line ahead ... can we do that in this case also when lines are not symmetrically separated?
Yes, it will work. It is customary to draw contour lines with an equal increment in potential between adjacent contours. It's true for topographic maps as well, if you're familiar with those.

aximwolf said:
Thank you tutors for your help that worked!
You're welcome. :smile:
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K