# Find the x and y component of an electric field

1. Dec 6, 2012

### warnexus

1. The problem statement, all variables and given/known data

http://i33.photobucket.com/albums/d86/warnexus/physicsfigure.jpg

2. Relevant equations
electric force = (field strength)(elementary charge)
electric force = k(e^2)/(distance^2)

3. The attempt at a solution
the goal of the problem is to find the x and y component of an electric field.
given distance 2.5*10^-2 m
elementary charge 1.6*10^-6 C
given k 9*10^9 N*m^2/C^2

we can find electric force (9*10^9)(1.6*10^-19)^2/(2.5*10^-2)^2 = 3.7 *10^-25 N

now we can find field strength.
field strength = electric force/(elementary charge)
field strength = (3.7*10^-25 N)/(1.6*10^-6 C)
the field strength came out to 2.3*10^-19 N/C

but i have no idea how to go about finding the components given the field strength. i need guidance
i'm pretty sure my works looks correct.

Last edited: Dec 6, 2012
2. Dec 6, 2012

### Staff: Mentor

You seem to be working only with single elementary charges, but there are two charges shown in the diagram which are not just single elementary charges. Both charges will produce their own electric field at a given location, and the total field is the sum of the two effects.

Also, you should be aware that you don't need to use the force expression, $F = k\frac{Q_1 Q_2}{r^2}$, but can directly employ the field expression, $E = k \frac{Q}{r^2}$ for each of the charges, and sum their individual contributions at a given point.

3. Dec 6, 2012

### ap123

It would be good for you to draw a diagram showing the directions of the 2 electric fields at the specified point. This will make finding the components easier.

4. Dec 7, 2012

### warnexus

I see so I am suppose use 2*10^-6 C as my charge. If I am solving for the left of P, that means my distance will half the given distance which is 2.5cm or 2*10^-2m, right?
given your input my field will be (9*10^9 N*m^2/C^2)(2*10^-6 C)/(2.5*10^-2 m)^2 = 2.88*10^7 N for the proton charge and -2.88*10^7 N for an electron charge.

Last edited: Dec 7, 2012
5. Dec 7, 2012

### warnexus

the diagram was given from my MasteringPhysics assignment.

6. Dec 7, 2012

### haruspex

For the positive charge, yes. What about the distance to the -ve charge?
Watch the units. The units in your answer are those of a force, not a field.
Distance is different, so can't be the same.
Having determined the magnitude of the net field, need to state its direction.

7. Dec 7, 2012

### Staff: Mentor

The idea is to annotate the given drawing to indicate things like the location for which you want to calculate the field, the distances from each charge to that point, and the directions and rough relative magnitudes of the fields that each charge will produce there; it doesn't have to be to scale, it's just to give you an idea of what direction you expect the net field to have and the signs to assign to the individual fields from each charge for when you sum them.

8. Dec 7, 2012

### Staff: Mentor

Yes, that would be the charge on the left. The charge on the right is -2*10^-6 C .
That's the distance of the location of interest to the nearest charge, the positive charge. You'll also need to know the distance from that location to the other, negative charge; since there are two charges you need to calculate the field contributions of each separately, then add them.
It would be better to refer to them as the positive and the negative charges since they are not individual elementary charges (they are much, much larger than that!). Also, electric fields are not specified in units of Newtons (at least not just Newtons). If you work through the units in your expression above you should find that the net units boil down to N/C (that is, Newtons per Coulomb). An equivalent set of units for the electric field is V/m (Volts per meter).

As to your numerical values, note also that you would not expect the field contributions from each charge to be the same magnitude since the location of interest is not situated at the same distance from both charges.

9. Dec 7, 2012

### warnexus

okay based on feedback, i think I have the idea. the electric field in the plane of the page 5.0cm to the left of P is by using the force expression (9*10^9)(2*10^-6)/(2*10^-2)^2 = 4.5 * 10^7 N/C.

i know i am suppose to use trig to find the x and y component but i dont think I have enough clue in the problem to figure out how to approach finding the components. i know the resultant is 4.5*10^7 N/C which the the electric field

http://i33.photobucket.com/albums/d86/warnexus/electricfield.jpg

10. Dec 7, 2012

### Staff: Mentor

Since the point of interest lies in the plane of the page, and so do the charges, and since the point and charges are co-linear along the assumed x-axis, what can you say about the y and z components of the field at the point? Remember that the electric field emanating from a point charge is radially directed.

Also, there are TWO charges to deal with here. Both will contribute to the net field.

I was expecting your diagram to end up looking something like this:

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Last edited: Dec 7, 2012
11. Dec 7, 2012

### haruspex

R2 = 7.5cm

12. Dec 7, 2012

### Staff: Mentor

Heh. I did say "something like this", after all...

Edit: Fixed the diagram.

Last edited: Dec 7, 2012
13. Dec 7, 2012

### warnexus

wait based on your diagram left of P is 5 cm. the y component looks smaller in magnitude than the x component from your diagram

14. Dec 7, 2012

### Staff: Mentor

What is shown is the directions and rough relative magnitudes) of the field vectors. E1 is due to the +2μC charge, and E2 is due to the -2μC charge. Yes they are different magnitudes! The charges are at different distances. There are no y-components because it is assumed that the charges and point are co-linear along the x-axis and the fields are emitted radially from the charges.

15. Dec 7, 2012

### warnexus

the x component turned out to be -2.6*10^7. man this question is hard. i must have spend 3hrs on this problem alone. physics hw sure consumes a lot of time.

16. Dec 7, 2012

### Staff: Mentor

Your result magnitude looks good. Be sure to always include units with results! You WILL lose marks if units are not specified.

Now that you understand the approach, the next dozen or so will go more quickly

17. Dec 8, 2012

### warnexus

i struggled trying to get the answer right. my online assignment gives me 5 attempts before it gives me the answer and I get no points for it. well at least i finished my physics hw assignment with a 84% =] cant believe i spend the whole day(must have been like 8 hrs) on the remaining 7 physics problems. im such a noob =/ at this rate how am i suppose to do well on the final? i feel when it comes to solving the vector component. i have no idea how to start and the text book does not give examples on how to go about this. physics is by far the most difficult class I ever taken and i thought calculus III was hard but solving physics problem right gives me a dopamine rush

18. Dec 8, 2012

### Staff: Mentor

The 'early days' of studying new material are always tough because one is working with new, incompletely understood concepts, few examples, and with little practice. But it does get better! After some practice and pondering on the concepts (like electric fields and potential, and how they behave around a charge), the ideas and methods will click and reach the "it's obvious" stage. Hopefully your previous mathematical studies will provide all the requisite tools; for the most part the math is just algebra (regular, linear, and vector) and some calculus.

Sometimes all you need is a few worked examples for similar problems in order to flesh out the concepts and procedures. Take advantage of Physics Forums' history! If you look at the bottom of the page inside a thread you'll see a section with the heading "Similar Threads for: ...." There you can follow links to see how others have faced and solved similar problems, along with the clarifying advice of others. Go browsing!

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19. Dec 9, 2012

### warnexus

thanks for the advice, gneill! =]