Solve part Abcd Problem 2199 Problem 2199 carries 300 HC of

Problem 21.99 Problem 21.99 carries 300 HC of charge dsributed untormly along Its length, and the other cames 300 pcdstributed uniformly along Las Mown in the K- 1.20 m 1,20 m Find the magniude of the electric feiseese wires poduce at point P. wrichis o am fom each wwe. Incorrect Try Again Part B thesowire produce a point P. AL4 Part C released Pwearis magnitude ofte nettoceeat these wives euwtonn Part D released at P wrat s Pe dreoton of eret force nat nere

Solution

a.

the electric field is calculated as follows:

    E = (kQ/d)[1 / sqrt[d² + (L/2)²]]

       = (9*10^9*3*10^-6/0.6)[1 / sqrt[0.6² + (1.2/2)²]]

       = 53033 N/C

Net electric field at point P is,

E = E₊ + E_-

     = -53033 N/C i^ - 53033 N/C j^

magnitude: E = 75000 N/C = 7.5e+4 N/C

b.

direction:

    @ = 90^o + tan-1[53033/53033], ccw from +yaxis

        = 135^o, counter clockwise from +y-axis, or 45^o

c.

the net force is,

   F = eE = (1.6X10^-19)(7.5e+4 N/C) = 1.2e-14 N = 12e-15 N

d.

direction: since the electron has negative sign, so the force is opposite to the field direction.

So, @ = 360 -45 = 315^o,    counter clockwise from +y-axis,