Electric Field of Equilateral Triangle Charge Distribution

[pinkyjoshi65]

three small, negatively charged spheres are located at the vertices of an equilateral triangle.The magnitudes of the charges are equal. Sketch the electical field in the region around this charge distribution, including the space inside the triangle.

ok..so all i know is that density of field lines will be more whr the field is greater..

 

[pinkyjoshi65]

um..ok..im tryin to work this out..so wht I did i place a positive test charge near each vertex. And since charge flows from th positive to the negative end, the field lines will be drawn from the positve to the negative direction.

 

[pinkyjoshi65]

will it be something lyke this: http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html

 

[pinkyjoshi65]

never mind..i got it..Could someone just check the solutions to these questions...
1) What is the distance between two protons experiencing a mutually repelling force of magnitude 4.0 x 10-11 N?

Fe=kq1q2/ r2
4*10-11= (9*109*1.602*10-19*1.602*10-19)/r2
r2= (9*1.602*1.602*10-19*10-19*109)/4*10-11
r2=5.7725*10-18
r= 2.40*10-9 m

2) Two point charges, +4.0 x 10-5 C and -1.8 x 10-5 C are placed 24 cm apart. What is the force on a third small charge of magnitude -2.5 x 10-6 C, if it is place on the line joining the other two midway between the originally given pair of charges?

q1= 4.0*10-3 C
q2= -1.8*10-5 C
q3= -2.5*10-6 C

Fq3q1= Kq3q1/r2
= 9*109*4*10-5*-2.5*10-6/0.12*0.12
= -90*10-2/0.12*0.12 = -6250*10-2
= -62.50 N (left)
Fq3q2= kq3q2/R2
= 9*109*-2.5*10-6*-1.8*10-5/0.12*0.12
= 2812.5*10-2
= 28.12 N (Left)
F= Fq3q1+Fq3q2
= -62.50+28.12= -34.38 N(left)

3) What are the magnitude and direction of the electric field strength at point Z in the situation below?
Negatively charged sphere “x” (-2.0 x 10-5 C) is left most, 60 cm to the right of it is a positively charged sphere “y” (8.0 x 10-6C). Point “z” is 30 cm to the right of “y”.

q1=-2.0*10-5 C
q2= 8.0*10-6 C

E1=kq1/r2
=9*109*-2*10-5/0.90*0.90
=-18*104/0.90*0.90
= -222200 N/C (Right)
E2= kq2/R2
=9*109*8*10-6/0.30*0.30
= 72*103/0.30*0.30
=800000 N/C (Left)
Hence E on Z= 800000-222200= 577800= -5.8*105 N/C (Left)

4) If a stationary charged test particle is free to move in an electrical field, in what direction will it begin to travel?
It would move in the direction of the field.

 

[snipez90]

just skimming over, the process looks good, but I would watch out for you directions. You won't trip up as easily if you use the MAGNITUDE of the charges when using coulomb's law (just getting the magnitude of the force) and then indicate the direction of the force based on the signs of the charges under consideration.

 

[pinkyjoshi65]

i did indicate the directions of the charges..Is question 4 fine?

 

[pinkyjoshi65]

Anyone?

 

[Shooting Star]

2) The forces are in same direction. Should be added.

3) Take one direction to be the positive one. Here, take it toward the right. E1 is -ve in value, E2 is +ve in value. You have written E2 as +ve to the left, which is incorrect. Final value is correct, since you've written -ve to the left, but a better way would be to fix the +ve dircn beforehand and adjust the sign.

4) Only when it starts. Afterward, the inertia will be there and the path will not be along the field.

 

[pinkyjoshi65]

2) but there is a negative sign. Hence it will result in a subtraction. Wont it?
3)But if I change the direction, then the direction of the answer will change too. It will become right.
3) After the field has started, which path will the charge take?

 

[Shooting Star]

2) The 3rd charge is -ve. The +ve charge will pull the 3rd charge, and the -ve one will push it away. Both forces are in the same dirn.

3) E2 is acting toward the right.

4) The motion will depend on the shape of the field.

 

[pinkyjoshi65]

2) ok, so the direction of all the forces will be right. But the final F will still involve subtraction, since Fq3q1 is negative. So the answer is right, on the direction has to be right.
3)so E1 is left, E2 is right, and E on Z is right. (left)

Also did you check the simulation out?

 

[Shooting Star]

2) Both the forces are to the left, so you have to add. If you consider right to be +ve, then Fq3q1 -ve and Fq2q1 is also -ve.

3) If you give proper directions to E1 and E2, the sign of E=E1+E2 will be automatically adjusted. You simply have to sum the two.

(I'll check out the simulation.)

 

[Shooting Star]

Yes, I've checked it out. You get a good idea of the electrcic field lines.

 

[pinkyjoshi65]

i am so confused...for Q2, if i take as the negative direction to be left. the Fq3q1 will b in the left direction and Fq3q2 will be in the right direction. But if i take into account the charges, then Fq3q1 will be to the right (attraction), and Fq3q2 will also be to the right. So the over all force will be to the right. But although we r adding both the forces, due to the negative sign, there will be a subtraction.

 

[pinkyjoshi65]

And for Q3, the direction of E1 is left, E2 is right, and E on Z ir right?

 

[Doc Al]

for Q2

i am so confused...for Q2, if i take as the negative direction to be left. the Fq3q1 will b in the left direction and Fq3q2 will be in the right direction. But if i take into account the charges, then Fq3q1 will be to the right (attraction), and Fq3q2 will also be to the right. So the over all force will be to the right. But although we r adding both the forces, due to the negative sign, there will be a subtraction.

Assuming the first two charges are placed such that the positive charge is 24 cm to the left of the negative charge, then:
Fq3q1 will point to the left (it's an attractive force)
Fq3q2 will also point to the left (it's a repulsive force)

So both contributions to the net force will be negative.

 

[pinkyjoshi65]

So Fq3q1 is -62.50 N (left), Fq3q2 is 28.12 N(left), and hence the net F is -62.50+28.12= -34.38 N (left). Is that right? And what about my posting on the other question..

 

[Shooting Star]

-62.50-28.12 = -90.62. The minus means to the left.

(Have patience till tomorrow.)

 

[Doc Al]

So Fq3q1 is -62.50 N (left), Fq3q2 is 28.12 N(left), and hence the net F is -62.50+28.12= -34.38 N (left). Is that right?

No, not right. To the left means negative, so both forces are negative:
Fq3q1 = -62.50 N
Fq3q2 = -28.12 N

 

[pinkyjoshi65]

ok..so even though we don't get a negative value when we calculate Fq3q2, it is negative because it is acting in the left direction. Is that right? So even in problems invoving charges, we have to take one direction as negative and one as positive..?

 

[Doc Al]

ok..so even though we don't get a negative value when we calculate Fq3q2, it is negative because it is acting in the left direction. Is that right? So even in problems invoving charges, we have to take one direction as negative and one as positive..?

Never just "plug numbers" into Coulomb's law and just use whatever sign comes out. (You can do that if you use the vector form of the law and a carefully defined coordinate system.) Instead, use the rule about opposite charges attracting and same charges repelling to figure out the direction of the force, and use the formula to find the magnitude of the force. Whether the force ends up being positive or negative depends on which way it points and the sign convention you are using.

 

[pinkyjoshi65]

ok..now i get it..so suppose a charge is negative, and I am to find the electri field, should I enter the charge as negative into the equation or enter it as a positive value and then give the sign depending upon the direction. In my book, even though the charge is negative, they sub in the value as positive into the equation, and then give a charge depending upon the direction.

 

[pinkyjoshi65]

so then for Q3, E1 is 222200N/C right, E2 is 800000N/C left. So E on Z will be -800000+222200= -5.8*10^5N/C or 5.8*10^5N/C Left. Sounds Right?

 

[Doc Al]

so then for Q3, E1 is 222200N/C right, E2 is 800000N/C left. So E on Z will be -800000+222200= -5.8*10^5N/C or 5.8*10^5N/C Left. Sounds Right?

Looks like you mixed up the directions. E1 is the field from the negative charge, thus it points left; E2 points right.

 

[pinkyjoshi65]

Y would E1 point left. The negative charge will be attracted to the positive charge. Won't it?

 

[Doc Al]

Y would E1 point left. The negative charge will be attracted to the positive charge. Won't it?

We're talking about the field from two charges, not the attraction between them. The field from a point charge points towards a negative charge and away from a positive charge.

 

[Shooting Star]

ok..so even though we don't get a negative value when we calculate Fq3q2, it is negative because it is acting in the left direction. Is that right? So even in problems invoving charges, we have to take one direction as negative and one as positive..?

(Previous post deleted because lots of posts in between, which I didn't realize.)

That's absolutely correct! Still, let's just go through it once more from scratch.

Part of your problem was that you were able to understand from the physics of the situation that the resultant force should point to the left, but when you are applying the formula for Coulomb’s law directly, you are a having a confusion in signs.

If q1 is at A and q3 is at C, then the vector force on q3 due to q1 is (kq1q3/r13^2)U13, where U13 is the unit vector in the direction of displacement AC.

If q2 is at B and q3 is at C, then the vector force on q3 due to q2 is (kq2q3/r23^2)U23, where U23 is the unit vector in the direction of displacement BC.

But U13 = -U23, since they point in opp directions.

Using this, if you now put in the values of the charges (including +ve and –ve signs), and add the forces, you’ll find that the ans comes as 90.62*(-U13), signifying that the force is in the opp direction of U13, i.e., in a dirn opp to AB.

 

[pinkyjoshi65]

umm..k..one last question, in my textbook, even though the charge is negative, they sub in the value as positive into the equation, and then give a charge depending upon the direction...Is that OK, or do we have to sub in the value as it is (positive or negative). If we do that then we get a negative in the answer..And then if that charge is in the left direction, then we give it a negative sign. (which is already present?)

 

[Shooting Star]

If you can understand in which direction the force is pointing, then you can just put in the value of the charge and adjust for the sign later. But electrostatic field is a vector field which has actually three components. If you sub in the actual values and treat the force as a vector, then later you'll understand quickly how to do the 3-d calculations.

 

[pinkyjoshi65]

umm...ok..so i'll choose the method of substituing the actual value say -2 into the equation and if its in the left direction,I leave the asnwer as it is..(since the answer will be negative)..Sounds ok?

 

[Shooting Star]

But you have to choose the vector carefully.

Force exerted on charge at B by charge at A is in the direction of AB, and electrostatic field intensity E at B due to charge at A is in the direction AB.

So, if the charge at A is actually negative, putting in the actual negative value gives you an intensity in the direction of -AB, as should be the case.

 

[pinkyjoshi65]

ok..thank you soo much...You have been a real help..:):)..really really appreciate it...I had a few questions abt Double and single slit equations..If you hve time, could you look into the thread entitled Double slit equations? Again thank You soo much..