- #1

linda300

- 61

- 3

Hey,

I've been working on this problem, it starts by asking for the potential, then from that the electric field, and finally it asks to calculate the work required to move a charge from infinite to the point it was originally.

It's a 2D problem so the electric field was E = (E

W = ∫F . dl

and F = qE = q (E

My question is, how do i integrate both x,y in the vector seperately from infinity to a or b?

Do I first do the only the Ex component and set y=b, then to work calculated from that add the work calculated using the Ey component and setting x=a ?

Or use parametrization ψ(x) = 1/t(a,b) 0<t<1 so infinity initially and (a,b) at t=1

then do ∫F(ψ) . -1/t^2(a,b) between 0 and 1,

I'm just unsure if your allowed to have a parametrization like that, which involve infinity since should be continuously differentiable. But in this case I want it to be infinity initially

I've been working on this problem, it starts by asking for the potential, then from that the electric field, and finally it asks to calculate the work required to move a charge from infinite to the point it was originally.

It's a 2D problem so the electric field was E = (E

_{x},E_{y}) and the location of the particle is at (a, b).W = ∫F . dl

and F = qE = q (E

_{x},E_{y}),My question is, how do i integrate both x,y in the vector seperately from infinity to a or b?

Do I first do the only the Ex component and set y=b, then to work calculated from that add the work calculated using the Ey component and setting x=a ?

Or use parametrization ψ(x) = 1/t(a,b) 0<t<1 so infinity initially and (a,b) at t=1

then do ∫F(ψ) . -1/t^2(a,b) between 0 and 1,

I'm just unsure if your allowed to have a parametrization like that, which involve infinity since should be continuously differentiable. But in this case I want it to be infinity initially

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