A point charge of -4 pC is located at x 4 m. y 2 m. A second point charge of 12

Question

A point charge of -4 pC is located at x 4 m. y 2 m. A second point charge of 12 AC is located at x -1 m, y 3 m (a Find the magnitude and direction of the electric field at x 1 m y 0 NCE Enter 12.5 points 0 attempt(s) made (maximum allowed for credit 35) [after that, multiply credit by 0.5 up to 10 attempts] (counterclockwise from plus x axis) Enter /2.5 points 0 attempt(s) made (maximum allowed for credit 5) [after that, multiply credit by 0.5 up to 10 attempts] (b) Calculate the magnitude and direction of the force on an electron at 1 m, y 0 N Enter 12.5 points 0 attempt(s) made (maximum allowed for credit 5) [after that, multiply credit by 0.5 up to 10 attempts] (counterclockwise from plus x axis) Enter 12.5 points 0 attempt(s) made (maximum allowed for credit 35) [after that, multiply credit by 0.5 up to 10 attempts]

Explanation / Answer

a) electric field is given by 9*10^9*Q/d^2

where Q=charge magnitude
d=distance between the charge and the point

the direction will be away from the charge and towards the point if the charge is positive.

direction will be towards the charge and away from the point if the charge is negative.

a)

electric field due to -4 uC:

as the charge is negative , direction will be towards the charge.
hence direction vector=(4,-2)-(-1,0)=(5,-2)

distance=sqrt(5^2+2^2)=5.3852 m

hence unit vector along the direction of the electric field=(5,-2)/5.3852=(0.92847,-0.3714)

magnitude of electric field=9*10^9*4*10^(-6)/5.3852^2=1241.4

hence in vector form , electric field=1241.4*(0.92847,-0.3714)=(1152.6,-461.04)

electric field due to 12 uC:

as the charge is positive , direction will be away from the charge.
hence direction vector=(-1,0)-(1,3)=(-2,-3)

distance=sqrt(2^2+3^2)=3.6056 m

hence unit vector along the direction of the electric field=(-2,-3)/3.6056=(-0.5547,-0.8321)

magnitude of electric field=9*10^9*12*10^(-6)/3.6056^2=8307.7

hence in vector form , electric field=8307.7*(-0.5547,-0.8321)=(-4608.3,-6912.8)

hence net electric field=(-3455.7,-7373.8) N/C

hence magnitude=sqrt(3455.7^2+7373.8^2)=8143.4 N/C

direction =244.9 degrees from +ve x axis, counterclockwise

b) as we know, force on any charged particle of charge q in an electric field of E is given by q*E.

here q=-1.6*10^(-19) C

==> force=charge*electric field=(5.5291,11.798)*10^(-16) N

magnitude=1.303*10^(-15) N/C

angle with +ve x axis=arctan(11.798/5.5291)=64.89 degrees

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