Solving Two Point Charges: Calculating Electric Field

[seanmcgowan]

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 ( F_electric= k_C(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:
F_electric= 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.

 

[LowlyPion]

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 ( F_electric= k_C(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:
F_electric= 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.

 

[seanmcgowan]

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.

 

[kbaumen]

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.

 

[Doc Al]

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.)

 

[seanmcgowan]

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?

 

[Doc Al]

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.

 

[seanmcgowan]

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?

 

[LowlyPion]

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?

 

[seanmcgowan]

um, yeah, i thought?

 

[Doc Al]

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

 

[seanmcgowan]

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

 

[LowlyPion]

um, yeah, i thought?

And what is Coulomb's constant here?

http://en.wikipedia.org/wiki/Coulomb's_law#Scalar_form

 

[seanmcgowan]

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

 

[seanmcgowan]

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

 

[Doc Al]

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.

 

[LowlyPion]

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?

 

[Doc Al]

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!)

 

[LowlyPion]

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

Come now. It's just an itty bitty sign.

 

[seanmcgowan]

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

 

[Doc Al]

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

There you go. Don't forget the units.

 

[seanmcgowan]

the answer would be 1.44*10^7

 

[seanmcgowan]

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