A question about electric field

[Cookiey]

Homework Statement
Two point charges a and b whose magnitudes are same positioned at a certain distance along the positive x-axis from each other. a is at the origin. The graph is drawn between electrical field strength and distance x from a. E is taken positive if it is along the line joining from a to b.

From the graph it can be decided that

1) a is +ve b is -ve
2) a and b are both +ve
3) a and b are both -ve
4) a is -ve and b is +ve
The attempt at a solution

As the electric field is defined with respect to a positive unit charge, a should be positive. This is because the field strength near a is tending to positive infinity, and according to the question, this is from a to b. So, this can be interpreted as repulsion.

As we move towards the middle of the graph, the strength decreases, almost to zero, so the field from point charge b must be cancelling it out. So b is also positive.

But why does the graph go towards positive infinity again? Because if b is also positive, won't it push our test positive unit charge away? This direction would be from b to a, right? Shouldn't the graph go to negative in that case?

The answer is given as a and b both being positive.

Thanks for helping!

 

[andrevdh]

How would one quantify the electric field in a region of space?
That is what procedure should be followed to measure its direction and magnitude at a point in space?
Or what is the definition of E?

 

[Cookiey]

How would one quantify the electric field in a region of space?
That is what procedure should be followed to measure its direction and magnitude at a point in space?
Or what is the definition of E?

The electric field at a point is defined as the electrostatic force per unit positive charge, and its direction is the same as the direction of the electrostatic force on the unit positive charge, right?

Im sorry, I don't understand my mistake. Which bit should I reconsider?

 

[ehild]

Homework Statement
Two point charges a and b whose magnitudes are same positioned at a certain distance along the positive x-axis from each other. a is at the origin. The graph is drawn between electrical field strength and distance x from a. E is taken positive if it is along the line joining from a to b.
View attachment 91749

From the graph it can be decided that

1) a is +ve b is -ve
2) a and b are both +ve
3) a and b are both -ve
4) a is -ve and b is +ve
The attempt at a solution

As the electric field is defined with respect to a positive unit charge, a should be positive. This is because the field strength near a is tending to positive infinity, and according to the question, this is from a to b. So, this can be interpreted as repulsion.

As we move towards the middle of the graph, the strength decreases, almost to zero, so the field from point charge b must be cancelling it out. So b is also positive.

But why does the graph go towards positive infinity again? Because if b is also positive, won't it push our test positive unit charge away? This direction would be from b to a, right? Shouldn't the graph go to negative in that case?

The answer is given as a and b both being positive.

Thanks for helping!

You are right, a must be positive and b too, as the electric field of the charges look to cancel at the middle of the distance.

 

[Cookiey]

You are right, a must be positive and b too, as the electric field of the charges look to cancel at the middle of the distance.

Thanks for replying!
So the graph given in the question is not possible with any combination of positive and negative charges?

 

[andrevdh]

No, I am just starting out from the basics.
A small positive test charge, charge q₊ say, is used to measure the electric field at a point in space.
It is then defined as E = F_e/q₊ where F_e is the force it experiences as a result of the field.
The direction of E is then in the same direction as F_e .
The graph seems to suggest a positive and negative direction for E, that is + towards +x and - towards -x.
This means as q₊ approaches charge b it is experiencing an ever increasing force towards the +x direction.
Which would suggest that b is negative as you thought.

 

[ehild]

Thanks for replying!
So the graph given in the question is not possible with any combination of positive and negative charges?

No. Maybe, they mixed the plot with that of the potential. If it was the electric field, it should be exactly zero at the center. The potential would have a minimum there.

 

[ehild]

The graph seems to suggest a positive and negative direction for E, that is + towards +x and - towards -x.
This means as q₊ approaches charge b it is experiencing an ever increasing force towards the +x direction.

It depends from what direction that q₊ charge approaches b. Moving from left towards b, it experiences force towards the -x direction.

 

[andrevdh]

Not stated where b is located though.
Should be where she drew it in pencil.
Okay, I get it - therefore the ve (volts) in the answers.

 

[Cookiey]

Not stated where b is located though.
Should be where she drew it in pencil.
Okay, I get it - therefore the ve (volts) in the answers.

Oh, the 've' is the 've' in posti've'. I'm sorry about that! My question sheet is a photocopy, there's a statement at the end of the original (which is cut off here) that says the graph is plotted between a and b only.

 

[andrevdh]

Then the graph probably just indicates that the E field is increasing and do not take the direction into account.

 

[ehild]

Then the graph probably just indicates that the E field is increasing and do not take the direction into account.

Then the graph probably just indicates that the E field is increasing and do not take the direction into account.

I think they plotted the potential function instead of the electric field. E should be zero somewhere on the line ab between the charges. It is not in the plot, there is a minimum instead. The potential goes to + infinity at the positive charges, and has a minimum between them.