Question

Electromagnetism: Point-like charges
Can I get some help on how to start this. My notes aren't really helping me right now.

Answer 1

What do u want to learn?

Answer 2

im trying to find the electric field at an arbitrary point.

Answer 3

q1=3uC, q2=2uC and q3=-3uC. they're located at R1=(0 0 0), R2=(0 3cm 0) and R3=(0 0 2cm) respectively. find the electric field at an arbitrary point R=(x y z) using the general result, calculate the field at the point (5cm, 5cm, 5cm)

Answer 4

so i guess step one i'll have to convert cm to m so that's R2=(0, 0.03m, 0) and R3=(0,0,0.02m) and the position of (0.05m, 0.05m, 0.05m)

Answer 5

gonna use this as well\[F=k_e \frac{ q_1q_2 }{ r^2 }\]

Answer 6

The arbitrary point 'R' is (x,y,z).

Distance of point 'R' from charge \(q_1\)=\(\sqrt{x^2+y^2+z^2}\)
Distance of point 'R' from charge \(q_2\)=\(\sqrt{x^2+(y-0.03)^2+z^2}\)
Distance of point 'R' from charge \(q_3\)=\(\sqrt{x^2+y^2+(z-0.02)^2}\)

Answer 7

I guess u r having problem in finding the resultant of the electric fields.

Answer 8

i think so. I need to refresh my memory on vectors and such. it's been a year since my last physics class.

Answer 9

this is the page from my notes. just trying to make sense of it.

Answer 10

@Irishboy123

he may help u

Answer 11

actually i learnt 3-d geometry 6 months ago

Answer 12

thank you for your help so far.

Answer 13

:)

Answer 14

so for the force of q1 on q2 i get 59.93N
and for q1 to q3 i'm getting -202.27N but i think that may be wrong... or all of this is wrong....

Answer 15

and the force of q3 on q2 is -41.62N

Answer 16

The resulting field will just be the individual forces from each charged particle. Since @jiteshmeghwal9 was so kind to do the distances for us the answer is just simply:

$$E_{tot} = E_1 + E_2 + E_3 = \frac{1}{4\pi\epsilon_0}\left[ \frac{q_1}{\sqrt{x^2+y^2+z^2}} + \frac{q_2}{\sqrt{x^2+(y-0.03)^2+z^2}}+ \\
\frac{q_3}{\sqrt{x^2+y^2+(z-0.02)^2}} \right]$$

To get the field in (5, 5, 5) all you have to do is put in x=5, y=5, z=5. Because this is expressed in m then convert the 5cm to m too.

Answer 17

Electric field is a vector quantity we cannot use simple algebraic methods to get the resultant

Answer 18

so it's the last part of my notes where it uses the sigma.

Answer 19

You can if you're only interested in the magnitude of the electric field as was implied by some of the answers of OP, but you are right - not in general. I can do the vector calcs a bit later.

Answer 20

so it's saying magnitude AND direction. so i guess we'll need to do some calculus. in my notes i think i see what i'm supposed to use.

Answer 21

this picture is a little closer but right there down at the bottom, i think that's what i'm supposed to use to solve this problem. do you agree?

Answer 22

or rather in the ovals.

Answer 23

ok nevermind i see what's up now.

Answer 24

that is what you need to do yes. The indices ij don't really "mean" all that much they just indicate that you have to do the sum by all pairs of charges and skip the rii situations (particle acting on itself, to avoid dividing by zero) because the distance then is zero.