HELP A question on electricity (electric field and electric potential)

[Kudo Shinichi]

HELP!A question on electricity (electric field and electric potential)

Homework Statement

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?

The Attempt at a Solution

a)E=KQ/r^2
E_total=(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
V_total=(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
V_total=(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
V_total=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]

E-Field is a vector field. Direction matters.

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

 

[Kudo Shinichi]

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]

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]

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]

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]

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]

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]

What angle is a) pointing?

Since the electric field is moving toward the negative charge, then it moves toward both points with 1 $$\mu$$C and -2.5
tan$$\theta$$=opp/adj=100000/83210.1
tan^-1(100000/83210.1)=50.2

 

[Kudo Shinichi]

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]

Since the electric field is moving toward the negative charge, then it moves toward both points with 1 $$\mu$$C 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]

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