Two d = 2.80-m nonconducting wires meet at a right angle. One segment carries +4

Question

Two d = 2.80-m nonconducting wires meet at a right angle. One segment carries +4.75 mu C of charge distributed uniformly along its length, and the other carries -4.75 mu C distributed uniformly along it, as shown in the Find the magnitude and direction of the electric field these wires produce at point P, which is 140.0 cm from each wire. (Assume the +x axis goes to the right.) magnitude N/C direction counterclockwise from the +x direction If an electron is released at P, what are the magnitude and direction of the net force that these wires exert on it? magnitude N direction counter clockwise from the +x direction.

Explanation / Answer

let d = 2.8 m

R = 140 cm = 1.4 m

charge per unit length, lamda = Q/d

= 4.75/2.8

= 1.696 micro C/m

Electric filed due to one rod

E1 = (lamda/2*pi*epsilon*R)*sin(theta)

= (2*k*lamda/a)*(d/2)/sqrt(d/2)^2 + R^2)

= (2*9*10^9*1.696*10^-6/1.4)*(1.4/sqrt(1.4^2 + 1.4^2) )

= 1.54*10^4 N/c

Enet = sqrt(E1^2 + E2^2)

= sqrt(2)*1.54*10^4

= 2.18*10^4 N/c <<<<--------Answer

direction : 225 degrees with +x axis <<<<--------Answer

b) F = q*Enet

= 1.6*10^-19*2.18*10^4

= 3.49*10^-15 N <<<<--------Answer

Direction : 45 degrees with +x axis <<<<--------Answer

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