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## Homework Statement

## Homework Equations

kq/r^2=E

λ=q/L

## The Attempt at a Solution

λ=8μC/1=8μC/m

dq=λdx

dE=k(dq)/(√(x^2+2^2))^2

dE=kλdx/(x^2+4)

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for x-direction

dE

_{x}=dEcosθ=(2kλdx)/(x^2+4)^(3/2)

E

_{x}

=2kλ∫dx/(x^2+4)^(3/2)

=2kx[x/4(x^2+4)^(1/2)] (from 0 to 1)

=2kλ/4√5=16099.68V/m

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for y-direction

dE

_{y}=-dEsinθ=-kλxdx/(x^2+4)^(3/2)

E

_{y}

=-kλ∫xdx/(x^2+4)^(1/2) (from 0 to 1)

=-kλ(-1/√5+0.5)=-3800.6V/m

------------------------------------------------------------

E

_{net}=√[(E

_{x})^2+(E

_{y})^2]=16.54kV/m

diretion=tanEy/Ex=-13.28degree

----------------------------------------------------------

but the answer of (b)(i) E=24.65KV/m diretion=-13.28degree

what is wrong during the calculation ?

also no idea on (b)(ii)

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