What is the potential difference between two points in an electric field?

[tetoo]

Homework Statement

A space probe travels towards a planet measuring the electrostaticpotential as it approches. the electric field strength is -500 N/C at the altitude of 600,000 KM and it decreases steadily until it reaches -400 N/C at 580,000 KM above the planet's surface . find the potential difference between these two points. help me pls

Homework Equations

The Attempt at a Solution

 

[SammyS]

Homework Statement

A space probe travels towards a planet measuring the electrostaticpotential as it approches. the electric field strength is -500 N/C at the altitude of 600,000 KM and it decreases steadily until it reaches -400 N/C at 580,000 KM above the planet's surface . find the potential difference between these two points. help me pls

Homework Equations

The Attempt at a Solution

Hello tetoo. Welcome to PF !

Where are you stuck ?

What have you tried ?

According to the Rules of this Forum, you must show an attempt at solving the problem before we can help you.

 

[tetoo]

i tried to get the total change in strength -500-400=-900N/C hten tried to change it into volts but i couldnt.
and thanks DR sammy

 

[SammyS]

i tried to get the total change in strength -500-400=-900N/C then tried to change it into volts but i couldn't.
and thanks DR sammy

Actually, the change in electric field strength is:

-500-(-400) = -100 N/C , but that's not relevant.​

You know two quantities for each of two situations. You know altitude and electric field strength. You should be able get two equations in two unknowns.

Hopefully, you know the formula for electric field strength at a distance, r, from the center of a sphere with total charge, Q.

Assume that the planet is spherical with radius, R. At some altitude, a, how far is the space probe from the center of the planet?

 

[gneill]

The difficulty with assuming a spherical charge on the planet as the source of the field is that the magnitude of said field appears to be decreasing as you approach the planet.

Instead, how about considering $d \vec{V} = -\vec{E} \cdot \vec{ds}$?

 

[tetoo]

i only studied electromagnetic induction and some electricity is there some law or formula to solve the problem i really appreciate your help i want to know under what topic is that thing

 

[gneill]

i only studied electromagnetic induction and some electricity is there some law or formula to solve the problem i really appreciate your help i want to know under what topic is that thing

The topic of field strength and potential difference for more general cases than simple point charges is usually covered in a calculus based course on fields (Gravitational as well as Electromagnetic). I think it was called "Wave and Modern Physics" when I took it.

In general the "formula" you'd apply would be to start with an expression for the electric field, say E(x), with respect to location x and then integrate -E(x)*dx from the starting point to the ending point.

The way this problem is set up you should be able to solve it without resorting to calculus; The problem states that the electric field strength "decreases steadily", so one might assume a linear relationship. If you make a graph E(x) versus x, then the required integral is the area under the curve (between the E(x) line and the x-axis).

 

[DrZoidberg]

I think you can simply work with the average field strength i.e. -450 V/m.

 

[gneill]

I think you can simply work with the average field strength i.e. -450 V/m.

Yes, that's true, given the linear form for the field strength.

 

[tetoo]

can anyone solve it in steps?

 

[pgardn]

The difficulty with assuming a spherical charge on the planet as the source of the field is that the magnitude of said field appears to be decreasing as you approach the planet.

Instead, how about considering $d \vec{V} = -\vec{E} \cdot \vec{ds}$?

So did you try the above?

The electric field became less negative in a linear manner as position changed...

Finding the area under a E. vs position graph is just multiplication if you use the avg. field strength.

 

[tetoo]

no i didnt get it i want the potential difference in volts no one wants to write the steps

 

[pgardn]

Actually I just gave you the path. It was already given above as well.

What is ds?

 

[tetoo]

what ds ?

 

[pgardn]

Instead, how about considering $d \vec{V} = -\vec{E} \cdot \vec{ds}$?

What is it?

 

[tetoo]

ask genill i don't know i didnt get anything from all you wrote i wanted the laws or formulas to get potential difference in volts and you are all talking about another thing ...

 

[pgardn]

ask genill i don't know i didnt get anything from all you wrote i wanted the laws or formulas to get potential difference in volts and you are all talking about another thing ...

Ok so what equation have you been given that is relevant to the problem?

 

[tetoo]

That's why i wrote the post Pgardn i don't know even the law... thanks for your help in advance

 

[pgardn]

We are not supposed to supply that.

What equations have you been using in class?

 

[tetoo]

this isnot a home work this is a sample question i suppose to search about its topic and find hte answer and study it I am in vacation and i will do test to apply for university unfortunayely this is out of book so i want some one to explain it in steps.

 

[DrZoidberg]

But you already got some good answers.
The equation you need is this
$d \vec{V} = -\vec{E} \cdot \vec{ds}$

If the field strength is constant you can simplify it to
$V = -E * s$

V is the voltage, E is the field strength and s the distance.
btw. 1 N/C is equal to 1 V/m

So all you really have to do is mutliply the average field strength with the distance.