Question

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.

Answer 1

E= 1.11 x 10^5 N/C

Step-by-step 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

`|OA| = \sqrt{0^(2) +3.10^(2) } \\|OA| =3.10 cm \\|OA| = 0.0310 m\\\\|BO| = \sqrt{4.71^(2) +0^(2) } \\|BO| =4.71 cm \\|BO| = 0.0471 m\\\\|OC| = \sqrt{7.50^(2) +8.70^(2) } \\|OC| =11.5  cm\\|OC| = 0.1149 m\\\\`

`E_(1) =(kQ_(1) )/(|OA|^(3) ) OA \\\\E_(1) =(9* 10^(9) * 52.1 * 10^(-9) )/(|0.0310|^(3)) (0.0000 i + 0.0310 j )\\\\E_(1) = (0.00 i + 487,929 j) N/C \\\\`

`E_(2) =(kQ_(2) )/(|BO|^(3) ) BO \\\\E_(2) =(9* 10^(9) * 25.4 * 10^(-9) )/(|0.0470|^(3)) (0.0470 i + 0.0000 j )\\\\E_(2) = (103486 i + 0 j ) N/C \\\\`

`E_(3) =(kQ_(3) )/(|OC|^(3) ) OC \\\\E_(3) =(9* 10^(9) * 18.2 * 10^(-9) )/(|0.1149^(3)) (0.0750 i - 0.0870 j )\\\\E_(3) = (8098.7 i - 9394.5 j ) N/C\\\\`

`E_(x) = E_(1x) +E_(2x) +E_(3x) \\E_(x) = (0+ 103486 +8098.7 )j\\E_(x)=111584.7 N/C\\E_(x) = 1.11 * 10^(5) N/C`