How Does Charge Placement Affect the Electric Field Calculation?

In summary: P should be 0, but it is not. i don't see how i can make the signs any other way. i followed the rule you gave me, and made sure i did not make the same mistake as before. do you mind telling me what is wrong with my understandingIn summary, the conversation discusses the calculation of the electric field at a point between two charges located 5cm apart. The conversation includes a diagram and equations for determining the electric field, as well as a discussion of the correct sign convention. It is
  • #1
jelliDollFace
26
0

Homework Statement



point A is between two charges, one to the left (+2 microcoul) and one to the right (-2 microcoul). point A is midway between the two charges. the two charges are 5cm apart. what is the electric field in the plane of the page (diagram) at a point 5cm left from point A. give in terms of x and y components of the electric field

assume positive x is directed to the right, and positive y is directed upward of point A

Homework Equations



electric field = kq/r^2 where k is 9*10^9, q is charge, r is distance

The Attempt at a Solution



E1 = kq_l/r_1^2
= (9*10^9)(+2*10^-6) / (0.025^2)
= +2.88*10^7 n/c

E2 = kq_r/r_2^2
= (9*10^9)(-2*10^-6) / (0.075^2)
= -3.2*10^6 n/c

Enet = E1 + E2 = (+2.88*10^7) + (-3.2*10^6) = +2.56*10^7 n/c

so Enet is directed to the right, purely x direction, no y direction

(Ex, Ey) = (+2.56*10^7, 0) n/c

my solution is incorrect. what did i do wrong? did i interpret the question correctly (plane part), if so could you explain what the "plane" part means?
 

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  • #2
jelliDollFace said:
point A is between two charges, one to the left (+2 microcoul) and one to the right (-2 microcoul).
Looks like you have the charges interchanged. (In your diagram and calculations, the positive charge is on the right.)
 
  • #3
no, the positive charge in the diagram is on the left, not the right. how would having the charge on either side affect the result?
 
  • #4
jelliDollFace said:
no, the positive charge in the diagram is on the left, not the right.
OK, I misread the diagram. But in your calculation, the charges are reversed.
how would having the charge on either side affect the result?
It changes the direction of the field.

I think the problem is that you are using the formula somewhat blindly to calculate the field and its sign. Instead, use the formula to get the magnitude of the field, then figure out the direction based on the sign of the charge. The field always points away from a positive charge and toward a negative charge.
 
  • #5
i can't see how i got the charges wrong in my calculations...q_l = q_left = left charge = positive charge = +2*10^-6 coulombs, and the distance is 2.5cm from the positive charge, so 0.025m.

as for the q_r = q_right = negative charge = -2*10^-6 coulombs, and the distance is 7.5cm from the negative charge, so 0.075m

am i not seeing something?
 
  • #6
jelliDollFace said:
i can't see how i got the charges wrong in my calculations...q_l = q_left = left charge = positive charge = +2*10^-6 coulombs, and the distance is 2.5cm from the positive charge, so 0.025m.
Which way does the field from that positive charge point? Should the sign of its field be positive or negative?

as for the q_r = q_right = negative charge = -2*10^-6 coulombs, and the distance is 7.5cm from the negative charge, so 0.075m
Which way does the field from that negative charge point? Should the sign of its field be positive or negative?
 
  • #7
the positive charge electric field points away from the charge, it should be negative

the negative charge electric field points into the charge, it should be positive

is i change the signs in my original calculations, will my answer be correct?
 
  • #8
That should fix it.
 
  • #9
thanks so much , it worked! it was the sign all along, that's something i need to remember
 
  • #10
just another quick question. if was to determine the electric field at point P, is this how i would do it.

E1 = kq_l/r_1^2
= (9*10^9)(-2*10^-6) / (0.025^2)
= -2.88*10^7 n/c

E2 = kq_r/r_2^2
= (9*10^9)(+2*10^-6) / (0.025^2)
= +2.88*10^7 n/c

so Enet = E1 + E2 = 0

so (Ex, Ey = 0, 0) correct? apparently it is wrong.

i know there is no field in the y direction so it has to be zero, and the distance from either charge is equal and only along the x axis, and since they are opposite directions, shouldn't they cancel out to zero?
 
  • #11
jelliDollFace said:
i know there is no field in the y direction so it has to be zero, and the distance from either charge is equal and only along the x axis, and since they are opposite directions, shouldn't they cancel out to zero?
Once again, check your signs. Which way does the field from the positive charge point? The negative charge?

(I assume you're talking about point A, right in the middle between the two charges.)
 
  • #12
i thought my signs were correct this time? i made them such that they were in the same directions as the first problem, after we corrected them. if i reverse the signs, the magnitude will be same, and wouldn't they just cancel out?

and yes i was talking about point A, in the middle

when is said 'opposite directions' i meant away from the positive, towards the negative. or is that incorrect wording? the positive points away towards the negative, and the electric field of the negative flows inwards to itself

so the positive charge, sign is negative, and the negative charge, the sign is positive. how would that not cancel out if you get same magnitude?
 
  • #13
jelliDollFace said:
i thought my signs were correct this time? i made them such that they were in the same directions as the first problem, after we corrected them. if i reverse the signs, the magnitude will be same, and wouldn't they just cancel out?
The situation is different when the point changes. (But the principle for figuring out the field direction remains the same, of course.)

when is said 'opposite directions' i meant away from the positive, towards the negative. or is that incorrect wording? the positive points away towards the negative, and the electric field of the negative flows inwards to itself
Actually, if you stuck to that last sentence you'd be OK. :wink:

so the positive charge, sign is negative, and the negative charge, the sign is positive.
While your previous sentence was OK, this conclusion is false. (Actually draw the field arrows on a diagram.)
 
  • #14
since the point is now between the charges, the positive points to the right so it is positive, and the negative charge's field is positive direction too
 
  • #15
jelliDollFace said:
since the point is now between the charges, the positive points to the right so it is positive, and the negative charge's field is positive direction too
Exactly. The fields don't cancel, they add up.

(If both charges where the same sign, then the field in the middle would be zero.)
 

1. What is an electric field on a plane?

An electric field on a plane is a vector field that represents the distribution of electric force on a flat surface. It is a fundamental concept in physics that helps explain the behavior of electrically charged particles.

2. How is an electric field on a plane calculated?

The electric field on a plane is calculated by dividing the electric force by the charge of the particle. This can be represented mathematically as E = F/q, where E is the electric field, F is the electric force, and q is the charge of the particle.

3. What is the direction of an electric field on a plane?

The direction of an electric field on a plane is always perpendicular to the surface of the plane. This means that the electric field lines are parallel to each other and point away from positively charged particles and towards negatively charged particles.

4. How does the distance from a charged particle affect the strength of the electric field on a plane?

The strength of the electric field on a plane decreases as the distance from a charged particle increases. This relationship is known as the inverse square law, which states that the electric field is inversely proportional to the distance squared between the charged particle and the plane.

5. What are some real-world applications of an electric field on a plane?

An electric field on a plane has many practical applications, such as in the design of electronic devices, the operation of electric motors and generators, and the development of technologies like solar cells and particle accelerators. It is also used in medical imaging techniques such as electrocardiograms and electroencephalograms.

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