Point Charges/Dipole - What is the Force/electric field/dipole moment/potential energ


by bmarson123
Tags: dipole moment, electric field, force, point charges, potential energy
bmarson123
bmarson123 is offline
#1
Aug21-11, 08:01 AM
P: 24
1. The problem statement, all variables and given/known data
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?


2. Relevant 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)

3. 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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quietrain
quietrain is offline
#2
Aug22-11, 12:42 AM
P: 651
Quote Quote by bmarson123 View Post
1. The problem statement, all variables and given/known data
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?


2. Relevant 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)

3. 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
quietrain
quietrain is offline
#3
Aug22-11, 01:14 AM
P: 651
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

quietrain
quietrain is offline
#4
Aug22-11, 01:32 AM
P: 651

Point Charges/Dipole - What is the Force/electric field/dipole moment/potential energ


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
quietrain
quietrain is offline
#5
Aug22-11, 01:47 AM
P: 651
iv) potential energy = kQ1Q2/ (distance between charges = 5) , again your magnitude is incorrect
bmarson123
bmarson123 is offline
#6
Aug22-11, 05:55 AM
P: 24
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?
bmarson123
bmarson123 is offline
#7
Aug22-11, 06:13 AM
P: 24
Quote Quote by quietrain View Post
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) ?
bmarson123
bmarson123 is offline
#8
Aug22-11, 06:28 AM
P: 24
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?
quietrain
quietrain is offline
#9
Aug22-11, 07:35 AM
P: 651
Quote Quote by bmarson123 View Post
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.
quietrain
quietrain is offline
#10
Aug22-11, 07:45 AM
P: 651
Quote Quote by bmarson123 View Post
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, thats 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)


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