Solving Two Point Charges: Calculating Electric Field

In summary: Do the same for the positive charge, which is to the right of the negative charge. In summary, the electric field at the midpoint between the two charges is negative.
  • #1
seanmcgowan
35
0

Homework Statement



Two point charges are 10.0 cm apart and have charges of 2.0 uC ( the u is supposed to be a greek symbol where the left side of the u is extended down) and -2.0 uC, respectively. What is the magnitude of the electric field at the midpoint between the two charges?

Homework Equations


Coulonmb's law ( Felectric= kC(q1*q2)/r^2


The Attempt at a Solution



I have absolutley no idea how to figure this one out. My textbook has left me high and dry on this one.

The only thng that I dd that makes any sense at all is:
Felectric= 8.99*10^-9(-4/100) = -3.6*10^-10

I need some serious help with this. I would appreciate it if someone showed me how to figure this thing out.
 
Physics news on Phys.org
  • #2


seanmcgowan said:

Homework Statement



Two point charges are 10.0 cm apart and have charges of 2.0 uC ( the u is supposed to be a greek symbol where the left side of the u is extended down) and -2.0 uC, respectively. What is the magnitude of the electric field at the midpoint between the two charges?

Homework Equations


Coulonmb's law ( Felectric= kC(q1*q2)/r^2

The Attempt at a Solution



I have absolutley no idea how to figure this one out. My textbook has left me high and dry on this one.

The only thng that I dd that makes any sense at all is:
Felectric= 8.99*10^-9(-4/100) = -3.6*10^-10

I need some serious help with this. I would appreciate it if someone showed me how to figure this thing out.

Draw a diagram. You will need to find the contribution from both charges at the point they ask. This is done easily enough through superposition. That is adding the results of considering the effect of one charge and then the other.

I would note that your distance 5 cm for each.
But the direction will be determined by whether the charge is + or -, and of course the direction your charge may be.
 
  • #3


Im sorry i still don't get it. Do I use 5 cm rather than 10? and what is superposition? i didnt quite get your first defenition. are you saying to only use q1 in one equation, then q2 in the other? I am sorry i really have a tough time with this stuff.
 
  • #4


What you are asked to find is electrical intensity in the midpoint between the two given charges, that is 5 cm from each of them. Do you know what is electrical intensity, how to calculate it and what is superposition? It's definitely explained in your textbook unless you're reading Kant's Critique of Pure Reason.

Electrical intensity (without calculus) is a force exerted on a charge over the charge placed in an electrical intensity, in other words - E=F/q or E=(k*q)/r^2. Remember that this is a vector and it's pointing to the charge if the charge is negative (for example an electron) and pointing away from the charge if it's positive (for example a proton. Now if there is more than one charge, than the net intensity at a certain point, is the sum of all intensity vectors at that point. Previous sentence describes the so-called superposition principle.

What you need to do is find the intensity of both charges at the particular point, add them and the resultant vector's magnitude is what you are looking for. But your book just couldn't have let you high dry on this one, because these are basics of electrostatics.
 
  • #5


seanmcgowan said:
Im sorry i still don't get it. Do I use 5 cm rather than 10?
You need to find the field from each charge at the point in question. And that point is 5 cm from each charge. (The formula you have in your first post is for the force between two charges. What you need for this problem is the field from a point charge.)
and what is superposition?
That just means to add up the field contributions from each charge.
i didnt quite get your first defenition. are you saying to only use q1 in one equation, then q2 in the other?
That's right. The field from q1 just depends on q1 and its distance to the midpoint; same for q2.

So, what's the magnitude and direction of the field from q1 at the midpoint? The field from q2? Add them up.

(Review kbaumen's post to learn the formula for calculating the field from a point charge.)
 
  • #6


I used Kc*(q/r^2) for both of the particles. unfortunately the two cancel each other out. am i supposed to make one of them positive? or am i using the worng equation?
 
  • #7


They don't cancel. The fields point in the same direction, so they add.

Do this: Pretend the negative charge is on the left. Find the field it creates at the midpoint. What direction would it point?

Do the same for the positive charge, which is to the right of the midpoint.
 
  • #8


ok so for the answer i would:

8.99*10^-9 *( 2/25)= 7.19*10^-10

so then I add the two and get: 1.45*10^-9 is that it?
 
  • #9


seanmcgowan said:
ok so for the answer i would:

8.99*10^-9 *( 2/25)= 7.19*10^-10

so then I add the two and get: 1.45*10^-9 is that it?

Distance is .05 m?
 
  • #10


um, yeah, i thought?
 
  • #11


seanmcgowan said:
ok so for the answer i would:

8.99*10^-9 *( 2/25)= 7.19*10^-10
Careful:

2 μC = 2*10^-6 C (μ means "micro")
5 cm = 0.05 m
 
  • #12


oh ok, so ill use the 2*10^-6 in place of the 2 then, and 0.05 instead of 5 ill let you know what i get
 
  • #14


ok the answer i got was 8*10^-4. then added it to itself and got 1.6*10^-3
 
  • #15


Coulomb's constant is 8.99*10^-9. at least, as far as my txt book is concerned.
 
  • #16


seanmcgowan said:
8.99*10^-9 *( 2/25)= 7.19*10^-10
Oops... I missed that error in Coulomb's constant, but LowlyPion caught it. See his link.
 
  • #17


seanmcgowan said:
ok the answer i got was 8*10^-4. then added it to itself and got 1.6*10^-3

109*10-6/10-4 yields a different magnitude doesn't it?
 
  • #18


seanmcgowan said:
Coulomb's constant is 8.99*10^-9. at least, as far as my txt book is concerned.
Double check that exponent. (It's only off by a billion billion!)
 
  • #19


Doc Al said:
Double check that exponent. (It's only off by a billion billion!)

Come now. It's just an itty bitty sign.
 
  • #20


oh woops, its 10^9 ok so then it would be... 1.44*10^7
 
  • #21


seanmcgowan said:
oh woops, its 10^9 ok so then it would be... 1.44*10^7
There you go. Don't forget the units.
 
  • #22


the answer would be 1.44*10^7
 
  • #23


thanks lonely pion. you and Doc Al helped me a LOT.
 

What is the formula for calculating electric field between two point charges?

The formula for calculating electric field between two point charges is E = kQ/r^2, where E is the electric field strength, k is the Coulomb's constant (8.99 x 10^9 N*m^2/C^2), Q is the magnitude of the charge, and r is the distance between the two charges.

Can the electric field between two point charges be negative?

Yes, the electric field between two point charges can be negative. This indicates that the direction of the electric field is opposite to the direction of the force on a positive test charge placed at that point.

How does the distance between two point charges affect the strength of the electric field?

The strength of the electric field between two point charges is inversely proportional to the square of the distance between them. This means that as the distance increases, the electric field strength decreases.

What units are used for electric field strength?

The SI unit for electric field strength is Newtons per Coulomb (N/C). However, it can also be expressed in Volts per meter (V/m).

Can the direction of the electric field between two point charges change?

Yes, the direction of the electric field between two point charges can change depending on the relative positions and magnitudes of the charges. It always points away from positive charges and towards negative charges.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
730
  • Introductory Physics Homework Help
Replies
3
Views
174
  • Introductory Physics Homework Help
Replies
28
Views
467
  • Introductory Physics Homework Help
Replies
4
Views
950
  • Introductory Physics Homework Help
2
Replies
68
Views
4K
  • Introductory Physics Homework Help
Replies
5
Views
635
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
32
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
748
  • Introductory Physics Homework Help
Replies
23
Views
1K
Back
Top