# Calculating Work and E-field from and equipotential contour graph

• aximwolf
In summary, the conversation is about calculating the work done by an external agent to move a charge and the magnitude of the electric field at a specific point. The equations used are W = q∆V and E = -∆V/∆L. The main question is how to interpret the graph to find the values for ∆V and ∆L, which can be done by counting the number of lines and determining the potential difference between them. The tutors also discuss the symmetry of the lines and the use of equal increments in potential for contour lines.
aximwolf

## Homework Statement

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.

The difference in elctric potential for two adjacent equi-potential "lines" is ±1 Volt.

aximwolf said:

## Homework Statement

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.

## What is an equipotential contour graph?

An equipotential contour graph is a visual representation of the electric potential at different points in space. It consists of lines, known as equipotential lines, that connect points with the same electric potential.

## How do you calculate work from an equipotential contour graph?

To calculate work from an equipotential contour graph, you can use the equation W = qΔV, where q is the charge and ΔV is the change in electric potential between two points. Simply multiply the charge by the difference in electric potential between the two points to find the work done.

## How can you determine the electric field from an equipotential contour graph?

The electric field can be determined by finding the slope of the equipotential lines on the graph. The electric field will be perpendicular to the equipotential lines and the stronger the field, the closer the equipotential lines will be to each other.

## What information can you gather from an equipotential contour graph?

An equipotential contour graph can provide information about the electric potential and electric field at different points in space. It can also show the direction and strength of the electric field, as well as the work done by the field on a charged particle.

## How does distance affect the electric potential and electric field in an equipotential contour graph?

In an equipotential contour graph, the distance between the equipotential lines represents the change in electric potential. The closer the lines are to each other, the greater the change in potential over that distance. As for the electric field, the closer the equipotential lines are, the stronger the electric field will be in that region.

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