Find Electric Field vector from Equapotential lines

[[V]]

Homework Statement

Determine the magnitude and direction of the electric field at point 1 in the figure.

Homework Equations

$$E=\frac{QK}{r^2}$$
$$V=-\frac{QK}{r}$$

The Attempt at a Solution

$$\int E dr=-V$$
When I take the derivative of both sides of this relation wrt to R, I get
$$E=\frac{-dv}{dr}$$

So, now I just plug in two points, using point 1 as my final (energy flows from high potential to low potential:
$$E=-\frac{100-50}{0.01}=-5000 E/m$$, so it is pointing down. However, the answer appears to be 3750, pointing down.

Can someone explain what I did wrong? Please use calculus terminology if possible!

Thank you!

 

[SammyS]

Actually if you use the dot product for vectors:

$$dV=-\vec{E}\cdot d\vec{r}$$

If vector dr is in a direction perpendicular to vector E, what is the result for dV ?

 

[[V]]

That would be zero correct? I'm not sure how this answers my problem though. Any charge that moves around the equapotential line will have moved perpendicular to an E vector, therefore would have preformed zero work, and ΔV=0

But how would I go about figuring out the E field at any particular point. Is my formula and derivation correct? ∫Edr=-dv, E=-dv/dr ?

What am I missing?

 

[Redbelly98]

The Attempt at a Solution

$$\int E dr=-V$$
When I take the derivative of both sides of this relation wrt to R, I get
$$E=\frac{-dv}{dr}$$

So, now I just plug in two points, using point 1 as my final (energy flows from high potential to low potential:
$$E=-\frac{100-50}{0.01}=-5000 E/m$$, so it is pointing down. However, the answer appears to be 3750, pointing down.

You could just as well use the points at 25V and 50V to come up with answer. However, it will be different than your calculation based on the 50V and 100V points. Try taking the average of the two results.

 

[[V]]

How would that work? This is not a uniform E field, I'm sure.

 

[[V]]

Up until point 1 it is uniform, then at 100v the slope (E) changes. How does taking the average of the two points give me the correct answer for point 1.

It worked, but can you please explain this in detail?

Thank you!

 

[Redbelly98]

You're welcome

Getting the E-field in this way is just an estimate, it's not a very precise method.

If you want a better "picture" of the situation, make a graph of V as you go along the central vertical line (the line containing points 1 and 2). The graph will have a series of points. You can then try to connect the points with a smooth curve -- this is an approximation to the actual potential, so the slope of the curve is an approximation to the actual E-field.

 

[SammyS]

Actually if you use the dot product for vectors:

$$dV=-\vec{E}\cdot d\vec{r}$$

If vector dr is in a direction perpendicular to vector E, what is the result for dV ?

That would be zero correct? I'm not sure how this answers my problem though. Any charge that moves around the equipotential line will have moved perpendicular to an E vector, therefore would have preformed zero work, and ΔV=0

But how would I go about figuring out the E field at any particular point. Is my formula and derivation correct? ∫Edr=-dv, E=-dv/dr ?

What am I missing?

Yes, the answer is zero.

This does give that the direction of the E field is perpendicular to the equipoential lines.

If vector dr is in the same direction as vector E, then dV is negative. That gives the in/out information.

Redbelly98 has you on the road to getting a good estimate for the magnitude.