Point Charges/Dipole  What is the Force/electric field/dipole moment/potential energby bmarson123 Tags: dipole moment, electric field, force, point charges, potential energy 

#1
Aug2111, 08:01 AM

P: 24

1. The problem statement, all variables and given/known data
Two point charges q_{1} = 2[itex]\mu[/itex]C and q_{2} = 2[itex]\mu[/itex]C are placed at r_{1} = (3,0,0) m and r_{2} = (0,0,4)m respectively i) What is the force of q_{1} (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 q_{1}q_{2} / (mod (r_{1}r_{2}) )^{3} ] (r_{1}  r_{2}) E_{T} = [itex]\sum[/itex] E_{i} E = (mod q_{i}) / 4[itex]\pi[/itex][itex]\epsilon[/itex] mod r_{i} p = qd U =Qq / 4[itex]\pi\epsilon[/itex] mod (r_{2}  r_{1}) 3. The attempt at a solution i) F = k [ 4x10^{9} / 5^{3} (3,0,4) = 0.288 (3,0,4) N ii) E_{1} = 2x10^{9} / 4[itex]\pi[/itex][itex]\epsilon[/itex]3 = 5.99454 i E_{2} = 2x10^{9} / 4[itex]\pi[/itex][itex]\epsilon[/itex]4 = 4.49590 k E_{T} = (6.0,0,4.5) N/C iii) d = r_{2}  r_{1} = (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! 



#2
Aug2211, 12:42 AM

P: 651

your ans seems weird microCoulomb is 10^{6} , so q1q2 = 4 x 10^{12} also, your formula of (r_{1}r_{2}) means r_{21} = force of q2 TO q1. which is not the question. the question wants force of q1 TO q2. so your formula should read r_{12} = r_{2}r_{1} doing so gives you r_{12}= (3,0,4) now ve q and +ve q gives you attraction = q_{1}q_{2} so when you put back into forumla, you get r_{12}=r_{21} thats how you get force of Q1 ON Q2, which is attractive, means it pulls Q2 towards Q1, given by r_{21} if you are still confused, read this , scroll down to vector form http://en.wikipedia.org/wiki/Coulomb's_law 



#3
Aug2211, 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 Efield from +Q1 is towards the ve x direction. at origin, the Efield 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 = kQ_{1} r_{1} / r_{1}^{3} + kQ_{2}r_{2} /r_{2}^{3} so k = 9x10^{9} Q_{1} = +2x10^{6} Q_{2} = 2x10^{6} r_{1}^{3} = 3^{3}=27 r_{1}=(3,0,0) r_{2}^{3} = 4^{3}=64 r_{2}=(0,0,4) resultant direction = r_{1}+r_{2}=(3,0,4) magnitude = (666, 0 , 281) Efield at origin = (3*666, 0 , 4*281) N/C 



#4
Aug2211, 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} 



#5
Aug2211, 01:47 AM

P: 651

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




#6
Aug2211, 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 q_{1} then surely it should be the force of q_{2} on q_{1}? Because otherwise it'll be the force q_{1} is excerting on q_{2}? I did look at the link but it just confused me more, could you explain it a bit more for me? 



#7
Aug2211, 06:13 AM

P: 24





#8
Aug2211, 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?




#9
Aug2211, 07:35 AM

P: 651

F = kQQ/r^{3} * (3,0,4) = 9x10^{9} * (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
Aug2211, 07:45 AM

P: 651

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 (xaxis) as per the diagram above, the Efield is pointing to the left. hence, your vector is (3,0,0) 


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