Recent content by eminem14

R is the vector pointing to the midpoint of the line charge from the origin. if the midpoint of the line charge is at (3.5, 3.5, 0) then R would be < 0 - 3.5, 0 - 3.5, 0 - 0 > ∴ R= -3.5 i -3.5 j Regarding Rmag-3 , the magnitude of R is a constant so it can be out of the integral thats how...

E= k∫p_L*dL*R/(R_mag^3). so that would mean (√2*k*ρ_L)/(R_mag^3) * ∫R dx ? 1. Is the upper limit of the integral L which is √(7^2 + 7^2) = 7√2 ? 2. Since R = -3.5i -3.5j that would mean: ∫R dx= -3.5∫dx -3.5 ∫(-dx) (Because dy=-dx) →∫R dx=0 that doesn't make any sense. Where's the mistake?

So the correct answer is which? And how would it work out using my method? (E(0,0,0) due to the linear charge as a vector?) could u show me the correct steps (just a modification of what i did)?

i know the correct e-field is -7.231 i - 7.231 j which is the result i got using the method i mentioned in the previous comment (integrating over L*(√2 / 2)). I am still confused regarding that solution if someone can clarify that for me I got dx+ dy because in the formula sheet provided by...

i got to a solution for the line charge: Q=ρ_L * dL= 4*10^-9 *(dx + dy) => k*ρ_L /_Rmagnitude^3∫Rx dx + k*ρ_LRmagnitude^3∫Ry dy where R is the vector pointing from the origin to the midpoint of the line charge => R= -3.5i -3.5j but what are the limits of the integral? i did it over the length...

Homework Statement A finite uniform linear charge ρ_L = 4 nC/m lies on the xy plane; start point and end point are (7,0,0) and (0,7,0) .While point charges of 8 nC each are located at (0, 1, 1) and (0, -1, 1). Find E at (0, 0 ,0) Homework Equations dE=ρ_L *dz'/4∏ε *...

Thats what i figured. In my calculations i changed the μC to C and assumed the voltage would be in volts. Hopefully that should be correct.

Thanks again and one final dumb question are the units for the voltage volts or mili volts?

Thanks for the help got the correct result. But I just want to understand something, why is it that the radii are irrelevant?

S is a constant though right?

The min and max for the shells are 1.0 and 1.5. So those r the limits of the integral but then where does s fall into the integral. Is it a constant multiplied by the integral?

Q would be (1/r2 * 4πr2dr)= 4∏ dr ∴ ∫kQ/r= 1/ε∫1/r dr for point A limits of the integral: 1.5 < r < s and ∫kQ/r= 10^-6/ε∫1/r dr for point B limits of the integral: 1.5 < r < s but how is s calculated for each point? is it the magnitude of position vector pointing to the point →...

so s>1.5 what would the integral be? isn't the volume of each shell would 4/3 * pi * (rout^3 - rin ^3)? and what about the coordinates of A & B? where do they fall into this if Q can be calculated would the integral be: k*Q*∫1/r^3 dr r between 1 and 1.5 ?

([10^-6]/(4*∏*ε)*∫∫∫\frac{sinθ*dr*dθ*d∅}{(r*sinθcos∅ - 3)^2 + (r*sinθ*sin∅ -4)^2 + (r*cosθ - 12)^2)^3/2} where the limits of the integral: 1<r<1.5 0 <= θ <= ∏ 0 <= ∅ <= 2∏ *the exponent of the denominator is 3/2 not 3

([10^-6]/(4*∏*ε)*∫∫∫\frac{sinθ*dr*dθ*d∅}{(r*sinθcos∅ - 3)^2 + (r*sinθ*sin∅ -4)^2 + (r*cosθ - 12)^2)^3/2} where the limits of the integral: 1. 1<r<1.5 2. 0 <= θ <= ∏ 3. 0 <= ∅ <= 2∏