Solving Two Point Charges: Electric Field, Dipole Moment & Potential Energy

In summary: to +ve with the direction pointing from the negative charge to the positive charge.iv) the potential energy of an electric dipole is the sum of its electric potential energy and its kinetic energy.
  • #1
bmarson123
24
0

Homework Statement


Two point charges q1 = 2[itex]\mu[/itex]C and q2 = -2[itex]\mu[/itex]C are placed at r1 = (3,0,0) m and r2 = (0,0,4)m respectively

i) What is the force of q1 (in vector form)?

ii) What is the electric field at the origin?

iii) What is the electric dipole moment of this arrangement (in vector form)?

iv) What is the potential energy contained in this arrangement of charges?


Homework Equations



F =[ k q1q2 / (mod (r1-r2) )3 ] (r1 - r2)

ET = [itex]\sum[/itex] Ei

E = (mod qi) / 4[itex]\pi[/itex][itex]\epsilon[/itex] mod ri

p = qd

U =Qq / 4[itex]\pi\epsilon[/itex] mod (r2 - r1)

The Attempt at a Solution



i) F = k [ -4x10-9 / 53 (3,0,-4)
= 0.288 (3,0,-4) N

ii) E1 = 2x10-9 / 4[itex]\pi[/itex][itex]\epsilon[/itex]3
= 5.99454 i

E2 = 2x10-9 / 4[itex]\pi[/itex][itex]\epsilon[/itex]4
= 4.49590 k

ET = (6.0,0,4.5) N/C

iii) d = r2 - r1
= (-3,0,4)

p =2x10-9 (-3,0,4)
= (-6x10-9 , 0, 8x10-9) C.m

iv) U = -4x10-9 /4[itex]\pi[/itex][itex]\epsilon[/itex] 5 = 7.2 J

Basically, I just wanted to know if this is all right because it comes up on every single exam and I don't want to think it's right for it not to be!
 
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  • #2


bmarson123 said:

Homework Statement


Two point charges q1 = 2[itex]\mu[/itex]C and q2 = -2[itex]\mu[/itex]C are placed at r1 = (3,0,0) m and r2 = (0,0,4)m respectively

i) What is the force of q1 (in vector form)?

ii) What is the electric field at the origin?

iii) What is the electric dipole moment of this arrangement (in vector form)?

iv) What is the potential energy contained in this arrangement of charges?

Homework Equations



F =[ k q1q2 / (mod (r1-r2) )3 ] (r1 - r2)

ET = [itex]\sum[/itex] Ei

E = (mod qi) / 4[itex]\pi[/itex][itex]\epsilon[/itex] mod ri

p = qd

U =Qq / 4[itex]\pi\epsilon[/itex] mod (r2 - r1)

The Attempt at a Solution



i) F = k [ -4x10-9 / 53 (3,0,-4)
= 0.288 (3,0,-4) N

ii) E1 = 2x10-9 / 4[itex]\pi[/itex][itex]\epsilon[/itex]3
= 5.99454 i

E2 = 2x10-9 / 4[itex]\pi[/itex][itex]\epsilon[/itex]4
= 4.49590 k

ET = (6.0,0,4.5) N/C

iii) d = r2 - r1
= (-3,0,4)

p =2x10-9 (-3,0,4)
= (-6x10-9 , 0, 8x10-9) C.m

iv) U = -4x10-9 /4[itex]\pi[/itex][itex]\epsilon[/itex] 5 = 7.2 J

Basically, I just wanted to know if this is all right because it comes up on every single exam and I don't want to think it's right for it not to be!

i) i assume question wants force OF q1 ON q2
your ans seems weird
microCoulomb is 10-6 , so q1q2 = -4 x 10-12

also, your formula of (r1-r2) means r21 = force of q2 TO q1. which is not the question. the question wants force of q1 TO q2. so your formula should read r12 = r2-r1

doing so gives you r12= (-3,0,4)

now -ve q and +ve q gives you attraction = -q1q2

so when you put back into forumla, you get -r12=r21

thats how you get force of Q1 ON Q2, which is attractive, means it pulls Q2 towards Q1, given by r21

if you are still confused, read this , scroll down to vector form
http://en.wikipedia.org/wiki/Coulomb's_law
 
  • #3


ii) the electric field at a point is the superposition of the point charges everywhere else

so question wants at origin.

at origin, the E-field from +Q1 is towards the -ve x direction.
at origin, the E-field from -Q2 is towards the +ve z direction.

so your resultant direction is not (6,0,4.5), i think your magnitude is wrong too

E = kQ1 r1 / r13 + -kQ2r2 /r23

so k = 9x109
Q1 = +2x10-6
Q2 = -2x10-6
r13 = 33=27
r1=(-3,0,0)
r23 = 43=64
r2=(0,0,4)

resultant direction = r1+r2=(-3,0,4)
magnitude = (666, 0 , 281)
E-field at origin = (-3*666, 0 , 4*281) N/C
 
Last edited:
  • #4


iii) the electric dipole's displace vector d points from -ve to +ve charge.

so d is not (-3,0,4) but rather (3,0,-4)

your q should be 2*10-6
 
  • #5


iv) potential energy = kQ1Q2/ (distance between charges = 5) , again your magnitude is incorrect
 
  • #6


Thanks so much for your help!

I'm still a bit confused about i)

I thought that if it wants the force on q1 then surely it should be the force of q2 on q1? Because otherwise it'll be the force q1 is excerting on q2?

I did look at the link but it just confused me more, could you explain it a bit more for me?
 
  • #7


quietrain said:
ii) the electric field at a point is the superposition of the point charges everywhere else

so question wants at origin.

at origin, the E-field from +Q1 is towards the -ve x direction.
at origin, the E-field from -Q2 is towards the +ve z direction.

so your resultant direction is not (6,0,4.5), i think your magnitude is wrong too

E = kQ1 r1 / r13 + -kQ2r2 /r23

so k = 9x109
Q1 = +2x10-6
Q2 = -2x10-6
r13 = 33=27
r1=(-3,0,0)
r23 = 43=64
r2=(0,0,4)

resultant direction = r1+r2=(-3,0,4)
magnitude = (666, 0 , 281)
E-field at origin = (-3*666, 0 , 4*281) N/C

Also, why does r1 suddenly become negative? In the question you're told it's (3,0,0)m, so why does it become (-3,0,0) ?
 
  • #8


Sorry, I get i) now. I've recalculated it and I've got (863.2, 0, -1151.0) N. Is that a more reasonable answer?
 
  • #9


bmarson123 said:
Sorry, I get i) now. I've recalculated it and I've got (863.2, 0, -1151.0) N. Is that a more reasonable answer?

for i)

F = kQQ/r3 * (3,0,-4)
= 9x109 * (4x10-12) / 125 * (3,0,-4)
=0.000288 * (3,0,-4) N

i don't think your answer is right since it is quite huge.
 
  • #10


bmarson123 said:
Also, why does r1 suddenly become negative? In the question you're told it's (3,0,0)m, so why does it become (-3,0,0) ?

because r1 is not really the r1 given in the question per say

in the question, r1 is the coordinate, that's why its (3,0,0) , pointing from origin to that coordinate.

in ii), it wants the vector electric field at the origin.

so you know that electric field points outwards from +ve charges

so if your charge were at (3,0,0) , then where would it point at the origin? (you should draw it out to understand better) anyway,

0 <----------------- 3 (x-axis)

as per the diagram above, the E-field is pointing to the left. hence, your vector is (-3,0,0)
 

1. What is the formula for calculating the electric field between two point charges?

The formula for calculating the electric field between two point charges is E = kQ/r^2, where k is the Coulomb's constant (9x10^9 Nm^2/C^2), Q is the magnitude of the charge in Coulombs, and r is the distance between the charges in meters.

2. How do you calculate the dipole moment of a system of two point charges?

To calculate the dipole moment of a system of two point charges, you can use the formula p = Qd, where Q is the magnitude of one of the charges in Coulombs and d is the distance between the two charges in meters. The direction of the dipole moment is from the negative charge to the positive charge.

3. What is the significance of the dipole moment in an electric field?

The dipole moment in an electric field represents the strength and direction of the electric field. It is a measure of the separation of positive and negative charges in a system and can be used to determine the potential energy of the system.

4. How does the potential energy of a system of two point charges change with distance?

The potential energy of a system of two point charges follows an inverse relationship with distance. This means that as the distance between the charges increases, the potential energy decreases. This is because the electric force between the charges decreases as they move farther apart.

5. Is the potential energy of a system of two point charges always positive?

No, the potential energy of a system of two point charges can be either positive or negative depending on the relative positions of the charges. If the charges have the same sign (both positive or both negative), the potential energy will be positive. If the charges have opposite signs, the potential energy will be negative.

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