Answer
E = 23.76 N/C
Work Step by Step
Formula for electric field due to half arc line charge = kλ/r x (2 x vector)
Vector depends on which quadrant the field is acting on (eg i-hat or j-hat).
λ = q / (2rPi/2)
E = qk/(Pi r^2) x (2 x vector)
Both fields acting down so : 2 x vector = -2 j-hat
E = qk/(Pi r^2) x (-2 j-hat -2 j-hat)
E = 4qk/(Pi r^2) (-j-hat)
E = (4 x 15 x 10^(-12) x 8.99 x 10^9)/(Pi x 0.085^2) (- j-hat)
E = 23.76 (- j-hat)
Derivation of formula:
dE = k (dq/r^2) r-hat
dq = λdx
dx = rd\theta
dq = λrd\theta
r-hat = cos\theta i-hat + sin\theta j-hat
dE = k (λrd\theta / r^2) (cos\theta i-hat + sin\theta j-hat)
dE = kλ/r (cos\theta d\theta i-hat + sin\theta d\theta j-hat)
E = ∫dE = kλ/r (∫ cos\theta d\theta i-hat + ∫ sin\theta d\theta j-hat)
limits for integration will vary based on the quadrant the eletric field (E) is in.
