Point charge Q1 = -1.0 nC is at origin and point charge Q2 = +1 nC is on the x a

Question

Point charge Q1 = -1.0 nC is at origin and point charge Q2 = +1 nC is on the x axis at(+1m, 0, 0). A third charge Q3 = -2 nC is on the y axis at (0, + 2m,0). OObservation point P is at (+1 m, +2 m, 0).

a. Find the Electric field E1 due to Q1 at P.

b. Find the Electric field E2 due to Q2 at P.

c. Find the electric field E3 due to Q3 at P.

Express your result either in ( , , ) or i,j,k notation.

d. Find the total Electric field due to all three charges at point P. Also find its magnitude and direction.

Explanation / Answer

r1= sqrt(1^2 +2^2) = 2.24m

= tan^-2(y/x)= tan^-1(2/1) = 63.43 deg

E1= kQ1/r1^2 = (9*10^9*1.0*10^-9)/(2.24^2) = 1.79 N/C ….towards Q1

E1x= E1cos = 1.79cos63.43 = -1.5 N/C i

E1y= E1sin = 1.79sin63.43 = - 1.0 N/C j

E1= -1.5 i - 1.0 j

E2= kQ2/r2^2 = (9*10^9*1.0*10^-9)/(1.0^2) = 2.25 N/C ….away from Q2

E2x= 0 N/C

E2y= 2.25 N/C j

E2 = 0 i + 2.25 N/C j

E3= kQ3/r3^2 = (9*10^9*1.0*10^-9)/(1.0^2) = -9 N/C ….away from Q2

E3x= -9 N/C i

E3y= 0 j

E3= -9 N/C I + 0 j

Enet = (-1.5 -9.0) i + ( 1.0 +9.0)j

Enet = (-10.5i + 10.0 j) N/C

|Enet| = sqrt(10.5^2 +10.0^2) = 14.5 N/C

direction = tan^-1( -10.0/10.5) = 43.6 above –ve X axis

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