Calculating Work and E-field from and equipotential contour graph

AI Thread Summary
To calculate the work done by an external agent in moving a charge from point i to b, the potential difference (∆V) between these points must be determined using the equipotential contour graph. The potential at point i is found to be between 0V and +5V, allowing for the calculation of ∆V. For the electric field magnitude at point k, the relationship E = -∆V/∆L can be applied once ∆V and the distance (∆L) are identified from the graph. It is noted that even if contour lines are not symmetrically spaced, the potential difference between adjacent lines is typically consistent. Understanding these principles allows for accurate calculations of work and electric field strength.
aximwolf
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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.
 
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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:
 
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