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View Full Version : HELP!!!A question on electricity (electric field and electric potential)

Kudo Shinichi
Feb11-09, 04:39 PM
1. The problem statement, all variables and given/known data
Three point charges are located at the corners of an equilateral triangle as shown.
http://s5.tinypic.com/2hdvalh.jpg
a) Find the magnitude and direction of the electric field at P
b) Find the magnitude and direction of the electric field at P(The potential at infinity is zero)
c) What is the electric potential energy of a particle with charge 4 µC placed at P?
d) If the mass of the particle is 7.0*10^-11kg, what minimum velocity muse it be given so that it can escape to infinity?

3. The attempt at a solution
a)E=KQ/r^2
Etotal=(9*10^9x1*10^-6)/0.3^2+(9*10^9x2*10^-6)/0.3^2+(9*10^9x-2.5*10^-6)/0.52^2=216549.06 point up because the top point has the strongest charge (I am not sure whether I got the direction right or not)
b)V=KQ/r
Vtotal=(9*10^9x1*10^-6)/0.3+(9*10^9x2*10^-6)/0.3+(9*10^9x-2.5*10^-6)/0.52=46678.85
c)V=KQ1Q2/r
Vtotal=(9*10^9x1*10^-6x4*10^-6)/0.3+(9*10^9x2*10^-6x4*10^-6)/0.3+(9*10^9x-2.5*10^-6x4*10^-6)/0.52=0.187
d)V=sqrt(2kQ/mr)
m=mass
Vtotal=sqrt((9*10^9x1*10^-6)/(7*10^-11x0.3))+sqrt((9*10^9x2*10^-6)/(7*10^-11x0.3))-sqrt((9*10^9x2.5*10^-6)/(7*10^-11x0.3))=35500654.39
For part d I don't really know how to solve for velocity after I got the voltage

Even though I got the answer for all of them, but I don't think it is the right way to approach the problem, because the answers are either too big or too small

Please help me with it. Thank you very much.
LowlyPion
Feb11-09, 04:59 PM
E-Field is a vector field. Direction matters.

You can't just add the x,y scalars all together.
Kudo Shinichi
Feb11-09, 05:15 PM
E-Field is a vector field. Direction matters.

You can't just add the x,y scalars all together.

Part a
Etotal=((9*10^9x1*10^-6)/0.3^2)i+((9*10^9x2*10^-6)/0.3^2)i+((9*10^9x-2.5*10^-6)/0.52^2)j=300000i-83210.1j
Sorry, but I am still not sure which direction is the P pointing to, since j>i so i assume that it is pointing upward.

Also, is there any mistake for the other three parts?
Thank you for your help
LowlyPion
Feb11-09, 05:36 PM
Part a
Etotal=((9*10^9x1*10^-6)/0.3^2)i+((9*10^9x2*10^-6)/0.3^2)i+((9*10^9x-2.5*10^-6)/0.52^2)j=300000i-83210.1j
Sorry, but I am still not sure which direction is the P pointing to, since j>i so i assume that it is pointing upward.

Also, is there any mistake for the other three parts?
Thank you for your help

That's better, but there is a mistake in the net component along the i direction. There is a 1μ C in one direction and a 2μ C charge creating an opposing E-field.

The direction is determined by the tan-1 of the j/i
Kudo Shinichi
Feb11-09, 05:42 PM
That's better, but there is a mistake in the net component along the i direction. There is a 1μ C in one direction and a 2μ C charge creating an opposing E-field.


Point with 2μC has the positive direction along x-axis, point with 1μC has negative direction along x-axis
therefore, (9*10^9x2*10^-6/0.3^2)-(9*10^9x1*10^-6/0.3^2)=99888.89i

therefore, the answer for part a is 99888.89i-83210.1j since i>j and point with 2μC is greater than point with 1μC, the direction of P should be pointing to the point with 2μC (right)
LowlyPion
Feb11-09, 05:49 PM
The charge on the right is positive.

The E-field from a positive charge is outward.

The E-field at p will be -i directed. (To the left.)

In the j direction the charge is negative, hence inward, meaning j is upward directed.

The angle is as I have suggested previously.
Kudo Shinichi
Feb11-09, 05:55 PM
The charge on the right is positive.

The E-field from a positive charge is outward.

The E-field at p will be -i directed. (To the left.)

In the j direction the charge is negative, hence inward, meaning j is upward directed.

The angle is as I have suggested previously.

Oops, I forgot that the electric field is always moving to the direction of the negative charge...

I am wondering did I do the rest parts (part b, c and d)correctly?
From the answer I got from part a I am pretty sure that I did part b and c correctly, but can you help me with part d? thank you
LowlyPion
Feb11-09, 06:01 PM
Oops, I forgot that the electric field is always moving to the direction of the negative charge...

I am wondering did I do the rest parts (part b, c and d)correctly?
From the answer I got from part a I am pretty sure that I did part b and c correctly, but can you help me with part d? thank you

What angle is a) pointing?
Kudo Shinichi
Feb11-09, 06:12 PM
What angle is a) pointing?

Since the electric field is moving toward the negative charge, then it moves toward both points with 1 \muC and -2.5
tan\theta=opp/adj=100000/83210.1
tan-1(100000/83210.1)=50.2
Kudo Shinichi
Feb11-09, 06:31 PM
did i get the angle in part a) as well as the answers in other parts right?
Sorry for keep asking you whether I got the answer right for other parts...However, it is the only part I got stuck...and I want to get it done as soon as possible...hope you can forgive my rudeness
LowlyPion
Feb11-09, 06:44 PM
Since the electric field is moving toward the negative charge, then it moves toward both points with 1 \muC and -2.5
tan\theta=opp/adj=100000/83210.1
tan-1(100000/83210.1)=50.2

That would be +50.2° with respect to the -x axis?

d) is the potential energy from c) converted to ½mv²

I'm not checking your math.
Kudo Shinichi
Feb11-09, 06:51 PM
That would be +50.2° with respect to the -x axis?

d) is the potential energy from c) converted to ½mv²

I'm not checking your math.

yes the angle is part a is +50.2° with respect to the -x axis