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

Answer:

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

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

$E_{1} =\frac{kQ_{1} }{|OA|^{3} } OA \\\\E_{1} =\frac{9\times 10^{9} \times 52.1 \times 10^{-9} }{|0.0310|^{3}} (0.0000 i + 0.0310 j )\\\\E_{1} = (0.00 i + 487,929 j) N/C$

$E_{2} =\frac{kQ_{2} }{|BO|^{3} } BO \\\\E_{2} =\frac{9\times 10^{9} \times 25.4 \times 10^{-9} }{|0.0470|^{3}} (0.0470 i + 0.0000 j )\\\\E_{2} = (103486 i + 0 j ) N/C$

$E_{3} =\frac{kQ_{3} }{|OC|^{3} } OC \\\\E_{3} =\frac{9\times 10^{9} \times 18.2 \times 10^{-9} }{|0.1149^{3}} (0.0750 i - 0.0870 j )\\\\E_{3} = (8098.7 i - 9394.5 j ) N/C$

$E_{x} = E_{1x} +E_{2x} +E_{3x} \\E_{x} = (0+ 103486 +8098.7 )j\\E_{x}=111584.7 N/C\\E_{x} = 1.11 \times 10^{5} N/C$