Question

In a rectangular coordinate system, a positive point charge q = 8.00 nC is placed at the point x = +0.150 m, y = 0, and an identical point charge is placed at x = -0.150 m, y = 0. Find the electric field at the following points.

a. Find the x and y components of the electric field at x = 0.150 m, y = -0.400 m. (Units N/C)

b. Find the magnitude of the electric field at x = 0.150 m, y = -0.400 m. (Units N/C)

c. Find the direction of the electric field at x = 0.150 m, y = -0.400 m. (Units degrees)

Answer 1

Electric field is given by:

E = kq/r^2

direction of Electric field will be towards -ve charge and away from +ve charge

q1 = q2 = 8.00 nC = 8.00*10^-9 C

Charge q1 is at (0.150, 0) m, and Charge q2 is at (-0.150, 0) m

We need electric field at point P (0.150, -0.400) m

Electric field due to q1 at point P will be away from the charge, means in negative y-axis (since both charge are located at line x = 0.150 m)

E1 = kq1/r1^2 (away from charge)

r1 = sqrt((0.150 - 0.150)^2 + (-0.400 - 0)^2) = 0.400 m

then, E1 = (9*10^9)*(8.00*10^-9)/0.400^2 = 450 N/C (In -ve y-axis)

So, E1 = E1x + E1y

E1 = (0 i - 450 j) N/C

Electric field due to q2 at point P will be away from charge at deg below -ve x-axis

where, = arctan (0.400/(0.150 - (-0.150))) = 53.13 deg below -ve x-axis (in 4th quadrant x > 0 and y < 0)

E2 = kq2/r2^2 (away from charge)

r2 = distance between q2 and P = sqrt((0.150 - (-0.150))^2 + (-0.400 - 0)^2) = 0.5 m

then, E2 = (9*10^9)*(8*10^-9)/0.5^2 = 288 N/C

So, E2 = E2x + E2y

E2 = E2*cos i + E2*sin j

E2 = 288*cos 53.13 deg i - 288*sin 53.13 deg j

E2 = (172.8 i - 230.4 j) N/C

Now net electric field will be:

Enet = E1 + E2

Enet = (0 i - 450 j) + (172.8 i - 230.4 j)

Enet = (0 + 172.8) i + (-450 - 230.4) j

Enet = (172.8 i - 680.4 j) N/C

Part (a)

x-component of net electric field = Ex_net = 172.8 N/C

y-component of net electric field = Ey_net = -680.4 N/C

Part (b)

Magnitude will be:

|E_net| = sqrt (172.8^2 + (-680.4)^2) = 702 N/C

|E_net| = 702 N/C

Since Ex > 0 and Ey > 0, So Electric field is in 1st quadrant

Now direction will be given by:

Direction = arctan (Ey/Ex) = arctan (-680.4/172.8) = -75.75 deg

Direction = 75.75 deg below +ve x-axis = 360 - 75.75 = 284.25 degrees Counterclockwise from the +x-axis

Let me know if you've any query.