Question

3 Problem #2 Two charges, Q1+1.00 x 10-9C; Q2,=-1.00 x 10-9C 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 origin of the coordinate system a. The distance between charge 1 and the origin. b. The distance between charge 2 and the origin. c. The Electric field, in polar notation, at the origin.

Answer 1

2.)

a.

Distance of charge 1 from origin(d1) = sqrt((0.00-0.00)^2 + (2.00-0.00)^2)

d1 = 2.00 m

b.

Distance of charge 2 from origin(d2) = sqrt((2.00-0.00)^2 + (0.00-0.00)^2)

d1 = 2.00 m

c.)

given Q1 = +1.00*10^-9 C

Q2 = -1.00*10^-9 C

Since, Electric field due to charge particle is given by,

E = k*q/r^2

where, k = 9*10^9

q = magnitude of charge

r = distance from charge particle

So, Electric field due to Q1,

E1 = k*Q1/d1^2

E1 = (9*10^9)*(1*10^-9)/(2)^2

E1 = 2.25 N/C towards -y axis.

Electric field due to Q2,

E2 = k*Q2/d2^2

E2 = (9*10^9)*(1*10^-9)/(2)^2

E2 = 2.25 N/C towards +x axis.

Total electric field at origin(Enet) = 2.25 i + 2.25 (-j)

magnitude of electric field(|Enet|) = sqrt(2.25^2 + 2.25^2)

|Enet| = 3.18 N/C

direction($\Theta$) = arctan(Ey/Ex)

$\Theta$ = arctan(-2.25/2.25)

$\Theta$ = 45 deg from +x axis in clockwise direction.

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