Electric Field at Point 1 cm Left of Middle Charge (-3.5 µC Force)

[BunDa4Th]

Homework Statement

(a) Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in Figure P15.10.

http://www.webassign.net/sf5/p15_10.gif

That hard to read number is 3 cm.

(b) If a charge of -3.50 µC is placed at this point, what are the magnitude and direction of the force on it?
Magnitude
N

Homework Equations

E = Keq/r^2

The Attempt at a Solution

I tried using the formula but was incorrect.

E = 8.99e9(1.5e-6)/.003^2

I used 1.5e-6 for q because i thought you need to use the middle charge and i think the r is .003 m since from -2uC to 1.5uC its 2 cm but it say its 1 cm to the left of the middle charge so i thought i add the distance.

For B i am not even sure how to do since i don't even understand how to do part a.

 

[robb_]

The other charges contribute to the field at that point as well.

 

[BunDa4Th]

so all three charge is contribute to the point?

if so do i have to find the magnitude of each charge and add or subtract them?

or can i just find the magnitude of 1.5uC and do E = F/q?

 

[robb_]

All charges contribute, remember the field permeates space.
The magnitude matters.

 

[BunDa4Th]

Okay i know the magnitude but I am not sure how i would go about solving it.
I used the formula F = k|q1||q2|/r^2 finding each magnitude with the other charges affecting one or the other some way.

6uC = 46.735 N

1.5uC = 157.2813 N

-2uC = 110.5463 N

But, i am not sure how i would go about finding the electric field strength.

 

[robb_]

6uC = 46.735 N

1.5uC = 157.2813 N

-2uC = 110.5463 N

This confuses me. [C] = [N] ?
The electric field formula you stated is correct. You need to use it for each charge, ignoring the others , and find the field at the point of interest. Then add them together to find the total. The field is great in that respect, you can just add em up; but remember that the field is a vector.

 

[BunDa4Th]

okay, i am lost now.

 

[BunDa4Th]

do i just find each individual like:

8.99e9 (6e-6)/.01^2 = 5.394e8

K(1.5e-6)/.01^2 = 1.349e8

k(-2e-6)/.01^2 = -1.798e8

 

[jackiefrost]

Think of the three charges as laying on the x-axis with the left charge q1 at the origin. The field at any point along the x-axis is the vector sum of the individual fields that each charge would produce by itself. This is called the principle of superposition. The fact that all charges lie along a straight line helps to simplify the situation.

(a). To calculate the electric field E produced by a set of i charges at some point P in space consider the following equation:

http://home.comcast.net/~perion_666/Files/Eq-1.gif

E is calculated for a point in space given by the point's position vector R_p. For your problem we are only dealing with some point P located "1 cm (0.01 meters) to the left of the middle charge". You have to watch your units! Try to express evrything in SI units (Meters, Coulombs, Newtons, etc). So you'll have to express all your distances as meters rather than cm and charge in Coulombs rather than uC. Epsilon sub 0 is the "permittivity of free space" constant which in SI units is about 8.85 x 10^-12 C^2/Nm^2

r_i is the distance (in meters) between each charge q_i and P. Vector u_i is the unit vector in the direction from the charge to P. Since each charge is positioned on the x axis, each unit vector will equal either plus or minus 1, depending on whether P is to left (minus) or right (plus) of the particular charge being evaluated.

So, if q1 is located at x=0, then your point P is at x=0.02 meters, q2 is at x=0.03 meters, and q3 is at x=0.05 meters. From these you can figure the distances r_i between each q and P and the correct signs for each of the unit vectors from each q to P. Furthermore, q1 and q2 are positive charges while q3 is negative so use those signs for each q in its calculation. Each E field strength vector is calculated for each charge of three charges and they are all added together: E(r) = E(1) + E(2) + E(3) where E(1) is the calculation using q1, etc.

(b). Now that you've calculated the field strength vector at P, find the force it would have on the -3.50 uC (-3.5 x 10^-6 Coulombs) test charge by multiplying the value of E at P by the value of the charge: i.e. F = qE . The resulting force is in Newtons if you kept all your units constant.

Remember, the Electric field is a value that gives the force per unit charge at a point in space without reference to any test charge actually existing at that point. If a charge is actually inserted there it's a matter of multiplying that charge value by the force per unit charge to arrive at the actual force exerted on the test charge.

Apologies if this is confusing and I hope I haven't stated something erroneously - someone will correct me, hopefully

 

[BunDa4Th]

Its not so confusing but the given formula you wrote is new to me, since i have not seen that formula before.

I will give this problem another try hopefully when i get home today or during the weekend.

Thanks for the help and explanation.

 

[robb_]

the formula that jackiefrost gave you is the just like the equation you wrote for E but for several point charges, that is "i" of them.

 

[jackiefrost]

I neglected to mention that the direction of your calculated electric field vector E at point P is the direction that a positive test charge would accelerate in if inserted at P.

In your problem, since all charges are on the x axis, the resultant E vector must also point along the x axis, directed either toward +x if positive, or toward -x if negative (or be zero). Therefore, if E is positive at P (points from P toward +x direction), a positive test charge placed at P would accelerate in +x direction, a negative test charge would accelerate in -x direction. If E at P is negative the opposite would occur. Of course, if E were zero at P there should (ideally) be no force exerted on either polarity test charge placed there. Also, your calculated polarity and magnitude for E is only valid at point P since we weren't required to find E as general function of any point on the x axis. That isn't really too much harder to do but your problem didn't need it.

Does this make sense? For a better picture maybe check out http://en.wikipedia.org/wiki/Electric_field"

BTW - I'm just studying these topics myself an am certainly no expert so hopefully others will correct any errors I have made.

Good luck!