Three point charges are located in the x-y plane. q1 = -52.1 nc at (0.00, +3.10 cm), q2 = +25.4 nc at (-4.70 cm, 0.00) and q3 = -18.2 nc at (+7.50 cm, -8.70 cm). determine the x component of the net electric field at the origin (0.00 m, 0.00m), due to the charges. give your answer in the form "[+/-a. bc x 10^(x) n/c]j". note that j is the unit vector in the y direction.

E= 1.11 x 10^5 N/C

Explanation:

Given

Magnitude of charges

Q1 = -52.1 nC

Q2 = +25.4 nC

Q3 = -18.2 nC

Location of point Charge Q1, A= (0.00 i + 3.10 j) cm  

Location of point charge Q2, B = (-4.71 i + 0.00 j) cm

Location of point charge Q3, C = (7.50 i -8.70 j) cm

location of the origin is O = (0.00 i + 0.00 j) cm

Solution

The eleectric feild will be directed away from the positive charge along the line BO and towards the negative charges along lines OA and OC respectively

OA = A -O = (0.00 i + 3.10 j) cm  = (0.0000 i + 0.0310 j ) m ( direction of the

BO = O - B = (4.71 i - 0.00 j) cm = (0.0471 i + 0.000 j ) m

OC = C -A = (7.50 i -8.70 j) cm = (0.0750 i + 0.0870 j ) m

3

step-by-step explanation:

10 - 2.8 = 7.2

2 2/5 = 2.4

7.2 / 2.4 = 3

answer; equivalent to a one(1) horsepower motor;