Question

Two charges. Q1 +1.00 Times 10^-9C; Q2 = -1.00 Times 10^-9 C are positioned as follows: Q1 at (0.00 m, 2.00 m) and Q2 at (2.00 m, 0.00 m) Determine the following at the point (1.00 m, 1.00 m): The distance between charge 1 and the point. The distance between charge 2 and the point. The Electric field, in polar notation, at the point.

Answer 1

given that:

Q1=10^(-9) C at (0,2)

Q2=-10^(-9) C at (2,0)

part a:

vector from point 1 to point (1,1)=(1,1)-(0,2)=(1,-1)

distance=sqrt(1^2+(-1)^2)=1.4142 m

part b:

vector from point 2 to point (1,1)=(1,1)-(2,0)=(-1,1)

distance =sqrt((-1)^2+1^2)=1.4142 m

part c:

if vector from the point where the charge exists to the point where electric field to be calculated is R,

then electric field due to any charge q is given as

E=k*q*R/(magnitude of R)^3

where R is in vector form

k=coloumb's constant=9*10^9

electric field due to Q1=9*10^9*Q1*(1,-1)/1.4142^3=3.18198*(1,-1) N/C

electric field due to Q2=9*10^9*Q2*(-1,1)/1.4142^3=-3.18198*(-1,1) N/C

hence total electric field=6.36396*(1,-1) N/C

magnitude of electric field=6.36396*sqrt(1^2+(-1)^2)=9 N/C

angle with +ve x axis=arctan(-1/1)=-45 degrees