/sci/ - Science & Math

a rough sketch...

Compute an integral.

Damn /sci/ seems to be deserted today, any idea how the integral might look like?

>can't do trig
>can't do calculus
>can't do genphysics

Come on, OP. Coulomb's law, charge is a function of the angle, set up an integral, solve it.

>>5668683

Is there current? If nothing is moving in the system then the EM force is zero.

Otherwise, isn't the arc 1/3 of a solenoid? Couldn't you just calculate it as if n=1/3?

>>5668725
It's 2 dimensional

>>5668725
Wrong
>>5668733
Also wrong, sort of.

Because it's an arc it will only affect the force in the x direction (based on that drawing being an x-y plane) because the y-components in the arc will cancel out.

OP needs to integrate F_x=C*dq*cos(Theta) and switch the dq to a function of theta.

>>5668735
Isn't that C supposed to be an E

>>5668683
dF=(dq*Q)/(4*pi*E0*r^2)

Q<- the particle
dq<- very small piece of the arc
pi<- 3.14
E0<- vacuum permitivity
r<- radius of your arc

let L=dq/dl be the liniar density of your arc

thus dq=dl*L;

rewrite F=(Q*dl*L)/(4*pi*E0*r^2)
One of the components of F will be 0; Fx=0 cause of the simmmetry
Fy=F*sin(c)
so you now have

Fy=[(Q*dl*L)/(4*pi*E0*r^2)]*sin(c)

lenght of a tiny piece of the arc dl=r*dc
Fy=(Q*r*dc*L)/(4*pi*E0*r^2)*sin(c)
Integrate over your angle : from 0 to bla bla
The only part you have to integrate is sin(c)*dc the rest of the terms are constant
You should end up with sth
{(Q*r*L)/(4*pi*E0*r^2)}*(-cos(0)+cos(whatever angle you have))
where L=the total amount of charge on the arc / the lenght of the arc

>>5668683
no reply? so you managed to understand?

>>5671073
thank you kind sir