A wire having a uniform linear charge density Lambda is bent

A wire having a uniform linear charge density Lambda is bent into the shape below\' the arc is a quarter of a circle with radius L. Using only variables Lambda, L and the Coulomb constant Lambda, find the information below: What is the electric potential (V) due to the wire at point P, at the origin. Find the electric field for both the arc and the line charge separately. Only worry about the y direction for the arc.

Solution

A) for arc part:

every point on arc is equidistant from origin, d = L

and V = kq/d

V1 = k (pi L lambda /2) / L = k pi lambda / 2

for straight part:

taking a dx part at x distance,

charge on this part, dq = lambda dx

dV = k dq / x = k (lambda dx)/ x

integrating,

V2= k lambda (ln(x) )   x is from L to 3L

V2 = k lambda [ln(3) - ln(1)] = 1.1 k lambda

V = V1 + V2

= k lambda ( pi/2 + 1.1) = 2.67 k lambda

b) for staringh part:

dE = k (lambda dx)/ x^2 (-i)

integrating,

E1 = - k lambda / x

x is from L to 3L

E1 = - k lambda [ 1/3L - 1/L] = 2k lambda / 3L (-i)

for arc part:

taking d@ width of arc at angle @.

charge = L d@ lambda

dE = ((k L lambda d@) / L^2 ) (-cos@ i -sin@j)

integrating,

E2 = (k lambda/ L ) ( -sin@i + cos j)

@ is from 0 to pi/2

E2 = - k lambda / L   ( i + j)

E = E1 + E2

= (-1.667 k lambda / L )i   + (-1.667 k lambda / L ) j