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

In summary, the conversation discusses a problem involving point charges and their electric fields and potentials. The expert summarizer provides a summary of the conversation, stating that there is a mistake in the net component along the i direction and that the direction of the electric field at P is determined by the tan-1 of j/i. They also mention that the electric field is always moving towards the direction of the negative charge. Finally, they confirm that the angle in part a is +50.2° with respect to the -x axis and that the velocity in part d is 35500654.39m/s.
  • #1
Kudo Shinichi
109
1
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
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.
 
Last edited:
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  • #2


E-Field is a vector field. Direction matters.

You can't just add the x,y scalars all together.
 
  • #3


LowlyPion said:
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
 
  • #4


Kudo Shinichi said:
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
 
  • #5


LowlyPion said:
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)
 
  • #6


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.
 
  • #7


LowlyPion said:
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
 
  • #8


Kudo Shinichi said:
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?
 
  • #9


LowlyPion said:
What angle is a) pointing?

Since the electric field is moving toward the negative charge, then it moves toward both points with 1 [tex]\mu[/tex]C and -2.5
tan[tex]\theta[/tex]=opp/adj=100000/83210.1
tan-1(100000/83210.1)=50.2
 
Last edited:
  • #10


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
 
  • #11


Kudo Shinichi said:
Since the electric field is moving toward the negative charge, then it moves toward both points with 1 [tex]\mu[/tex]C and -2.5
tan[tex]\theta[/tex]=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.
 
  • #12


LowlyPion said:
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
 

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

1. What is an electric field?

An electric field is a physical field that surrounds an electrically charged particle or object. It is responsible for the force that acts on other charged particles or objects within its range.

2. How is an electric field created?

An electric field is created by the presence of a charged particle or object. The strength of the field depends on the magnitude of the charge and the distance from the charged object.

3. What is electric potential?

Electric potential is a measure of the potential energy per unit charge at a given point in an electric field. It is a scalar quantity and is often referred to as voltage.

4. How are electric field and electric potential related?

Electric field and electric potential are closely related. The electric field is the negative gradient of the electric potential, meaning that the direction of the electric field points in the direction of decreasing electric potential.

5. How can electric field and electric potential be measured?

Electric field and electric potential can be measured using various instruments such as electric field meters and voltmeters. These instruments can measure the strength and direction of the electric field as well as the difference in electric potential between two points.

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