Heya Help with Electro magnitisim Question Please

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The discussion revolves around calculating the total force acting on a +3 nC charge due to two +2 nC charges and determining the electric field at the position of the +3 nC charge. The user initially struggles with vector addition of forces and receives guidance on using the correct formulas and understanding vector components. After some back-and-forth, the user successfully calculates the net force as 9.536 x 10^-9 Xhat. The conversation shifts to another question about the deflection of an electron beam and the forces between two wires carrying current, prompting advice to start a new thread for further assistance. Overall, the discussion highlights the importance of vector analysis in electrostatics and the collaborative nature of problem-solving in physics.
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Heya everyone could u help me with these questions please



a charge +2nC is located at (0, 2 m). Another +2 nC charge is
located at (0, -2 m). A third charge, +3 nC, is located at (2 m, 0).
Calculate the total force (in vector form) acting on charge +3nC
from both +2nC charges.

Using the result obtained in part above, calculate the electric field at
the position of the + 3nC charge due to both +2nC charges.

please guys thanks alot
 
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vip_uae said:
a charge +2nC is located at (0, 2 m). Another +2 nC charge is
located at (0, -2 m). A third charge, +3 nC, is located at (2 m, 0).
Calculate the total force (in vector form) acting on charge +3nC
from both +2nC charges.

Using the result obtained in part above, calculate the electric field at
the position of the + 3nC charge due to both +2nC charges.

Hi vip_uae! :smile:

Show us what you've tried, and where you're stuck, and then we'll know how to help you. :smile:
 
well i used the normal formula of F13 = (Q1 . Q3)/ 4 pi E0 r13 multiply by Rhat 13 ... and i knw wats rhat13 and r13 that's for the first part of the question i add the F13 AND F23 to give me FTOTAL... is that right?
 
vip_uae said:
well i used the normal formula of F13 = (Q1 . Q3)/ 4 pi E0 r13 multiply by Rhat 13 ... and i knw wats rhat13 and r13 that's for the first part of the question i add the F13 AND F23 to give me FTOTAL... is that right?

Well, it looks ok …

but it would be more convincing if you showed us some figures. :wink:

erm … since you've got that far, what did you want help with? :smile:
 
its just i am getting a weired answer for the total force i am getting Zero in the end of it...

when i add F13 AND F23 = 0

and i don't understand what to they mean by VECTOR form
 
vip_uae said:
its just i am getting a weired answer for the total force i am getting Zero in the end of it...

when i add F13 AND F23 = 0

and i don't understand what to they mean by VECTOR form

Vector form means as a multiple of rhat, which you've done.

I don't understand how your F13 and F23 can add to zero, if you're adding them as vectors. :confused:

F13 is a multiple of the unit vector rhat13, while F23 is a multiple of rhat23.

Since rhat13 and rhat23 are not parallel, how can any combination of them add to zero?
 
awwwwww right sorry i actually missed out one part of it but it gave me a really small number for Xhat 9.537x10^-9

i just found somthing that if i want to have to calculate it in a vector form

F13 = Q1 X Q3 / 4 pi E0 (r13)^2 multiply by Rhat13 + (u0 X Q1 X Q3)/ 4 pi (r13)^2 Multiply [V3 \Delta(V1 \Delta Rhat13 )]

is that formula looks right it said that if a vector form needed Fe which is Electric Force + Magnetic force = the force
 
Last edited:
vip_uae said:
(u0 X Q1 X Q3)/ 4 pi (r13)^2 Multiply [V3 \Delta(V1 \Delta Rhat13 )]

Is that supposed to be a magnetic force?

Magnetic forces only act on moving charges.

These charges are stationary … so forget magnetic forces!

You're vector-adding a multiple of rhat13 to the same multiple of rhat23.

Use components … what is the resultant? :smile:
 
see i don't understand what you meant by Multiple of Rhat i knw r hat = r13/moduls r13
 
  • #10
vip_uae said:
see i don't understand what you meant by Multiple of Rhat i knw r hat = r13/moduls r13

Yes, you're multiplying the vector rhat13 by (Q1 . Q3)/ 4 pi E0 r13,

and adding to it the vector rhat23 multiplied by (Q2 . Q3)/ 4 pi E0 r23 …

aren't you? :confused:
 
  • #11
yes that's wat i am doing exactly
 
  • #12
vip_uae said:
yes that's wat i am doing exactly

ok, so you've got a sloping-down vector plus an equal sloping-up vector …

so their sum (resultant) is … ? :smile:
 
  • #13
ya i have got Fnet = 9.536 x 10^ -9 Xhat ... does this answer looks right?
 
  • #14
vip_uae said:
ya i have got Fnet = 9.536 x 10^ -9 Xhat ... does this answer looks right?

I haven't done the figures, but yes it certainly is a multiple of Xhat. :smile:
 
  • #15
Thats gr8 thanks a lot i have an another question please

when an electron beam shoots through a horizantal line in the Z direction carying 50 A .. how would the beam will deflect ... like the direction of the deflection how would it be :)
 
  • #16
ok, but same thing … show us how far you've got!

what sort of field is produced? which direction is it in? what is the equation that relates field strength and force? :smile:
 
  • #17
Hey tiny tim lol i found the answer some where basicly it will deflect upwards due to the way that the current is rotating ... soo o i got it straight away
 
  • #18
vip_uae said:
Hey tiny tim lol i found the answer some where basicly it will deflect upwards due to the way that the current is rotating ... soo o i got it straight away

he-he … now you see one of the reasons why we like to see some work first! :biggrin:
 
  • #19
hahahahahaha ohhhhhhhh i have a problem tooo :P see i have one of the questions asking me about 2 wires carying 10K A in different directions i need to see the force of one wire exerting on the other using the Motor Equation which i knw

F = L ( I upside down triangle B)

I is current L is length and B is magnetic Flux density

B = u H

H = I / 2 pi R *I guess*

should i replace each unit and see wat i can get?
 
  • #20
oooh … I always get confused by questions about induction. :confused:

Can you start a new thread on this one, so that some other people will answer? :smile:
 
  • #21
Awwwww lol its cool i submited ma assignment lol it was a big massive hastle hehehee
 
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