Question

At a particular moment, three charged particles are located as shown in the figure below. Q1 = −4.5 μC, Q2 = +5.5 μC, and Q3 = −7.0 μC. Your answers to the following questions should be vectors. (Recall that 1 μC = 1 ✕ 10−6 C. Assume that the +x axis is to the right, the +y axis is up along the page and the +z axis points into the page. Express your answers in vector form.) (a) Find the electric field at the location of Q3, due to Q1. E1 = −2.5e7 Incorrect: Your answer is incorrect. N/C (b) Find the electric field at the location of Q3, due to Q2. E2 = N/C (c) Find the net electric field at the location of Q3. Enet = N/C (d) Find the net force on Q3. Fnet,3 = N (e) Find the electric field at location A due to Q1. E1 = N/C (f) Find the electric field at location A due to Q2. E2 = N/C (g) Find the electric field at location A due to Q3. E3 = N/C (h) What is the net electric field at location A? EA = N/C (i) If a particle with charge −6.0 nC were placed at location A, what would be the force on this particle? Fon A =

Answer 1

Given

charges are Q1 = −4.5 μC, Q2 = +5.5 μC, and Q3 = −7.0 μC

(a) Find the electric field at the location of Q3, due to Q1

we knwo that the electric field due a point charge q at a distance r is

E = kq/r^2

so here the electric field at Q3 due to Q1 is , first q1 = -4.5*10^-6 C and the distance r = sqrt(3^2+4^2)cm = 5 cm = 0.05 m

E1 = 9*10^9*(-4.5*10^-6)/(0.05)^2 N/C = -16200000 N/C

(b) Find the electric field at the location of Q3, due to Q2.

Q2 = +5.5 μC , r2 = 3 cm = 0.03 m

E2 = 9*10^9*(5.5*10^-6)/(0.03)^2 N/C = 55000000 N/C

(c) Find the net electric field at the location of Q3

writing the components of the elelctric fields of E1,E2

E1 = E1x i + E1y j

E2 = E2x i + E2y j

E1 = E1 cos (135) i + E1 sin (135) j = 16200000 cos 135 i +16200000 sin 135 j

E1 = -11455129.85 N/C i + 11455129.855 N/C j

E2 = E2 cos90 i + E2 sin 90 j = 0 N/C i + 55000000 N/C

the net field is  

E3 = (-11455129.85 +0 ) i +(11455129.855 +55000000 ) j

E3 = - 11455129.85 i + 66455129.855 j

the magnitude is

E3 = sqrt((-11455129.85)^2+(66455129.855 )^2) N/C = 67435185.80 N/C

d) Find the net force on Q3.

Force F = kq1*q2/r^2

F1 is force due to q1 and F2 is force due to q2

F1 = kq1*q3/r^2 = 9*10^9*(4.5*10^-6*7*10^-6)/(0.05)^2 N = 113.4 N

the components are  

F1 = F1 cos (-45) i + F1 sin (-45) j

F1 = 113.4 cos(-45) i + 113.4 sin (-45) j

F1 = 80.186 N i - 80.186 N j

and  

F2 is force due to q2

F2 = kq2*q3/r^2 = 9*10^9*(5.5*10^-6*7*10^-6)/(0.03)^2 N =385 N

the components are  

F1 = F1 cos (90) i + F1 sin (90) j

F1 =385 cos(90) i + 385 sin (90) j

F1 = 0 N i + 385 N j

total force is F = (80.186+0) N i +(- 80.186 N +385 N ) j

F = 80.186 N i + 304.814 N j

the magnitude is F = sqrt(80.186^2 + 304.814^2) N

F = 315.185 N

e)

E_A = kq1/r^2 = (9*10^9*4.5*10^-6)/(0.03^2) N/C = 4.5*10^7 N/C

f) E_A = kq2/r2^2 = 9*10^9*(5.5*10^-6)/(0.05)^2 N/C = 1.98*10^7 N/C