Question

Page 2 September 14, 2018 Physics 1220 Examination 1 Fall 2018 Name (Last, First): 1. Consider two charges: q1 +2 μC and q: +3 μC. qi is at (-2 m, 0 m); q2 is at (+1 m, 0 m). a. (1 point) Show the two charges on the Cartesian Coordinate System provided. (5 points) Find the point on the x-axis where the total Electric Field produced by the two charges is zero. (Hint: Pick a point and express the distance from one charge to the point as "" and the distance from the other charge to the point as the separation of the charges minus ".) b. (4 points) Find the total Electric Potential at the point (0 m, +2 m) c. d. (5 points) Now imagine that a third charge, q 1 uC is placed at (0 m, +2 m). Find the net force (both magnitude and direction) on qa produced by qi and q2

professor says r= 1.35 and P is at (-0.65, 0).
He gave zero examples of worked through problems and the simple math is confusing me

Answer 1

Hi, you have encircled the sub-question (b) therefore, I guess you need answer to sub-question (b) only. Also, you have written specifically about this sub-question, " Professor says r = 1.35 ...., etc." Here, follows answer to sub-question (b)

Let q1 = charge at point (-2 m, 0 m) = +2 μC = +2 x 10^-6 C

Let q2 = charge at point (1 m, 0 m) = +3 μC = +3 x 10^-6 C

Let P = point on x axis between, q1 and q2, where electric field E = 0 V/m

Let E1 = electric field at point P due to charge q1

Let E2 = electric field at point P due to charge q2

Let k = Coulomb's constant = 9 x 10⁹ N.m²/C²

Let d = distance between q1 and q2 = 3 m

Let r = distance between q1 and P

Let (3 - r) = distance between q2 and P

E1 is given by:

E1 = k.q1/r² = k x 2 x 10^-6 C / r²

E1 = k x 2 x 10^-6 C / r² ---------------------(Eqn 1)

E2 is given by:

E2 = k.q2/r² = k x 3 x 10^-6 C / (3-r)²

E2 = k x 3 x 10^-6 C / (3-r)² -------------------(Eqn 2)

At P electric field net electric field E is zero. It means E1 and E2 are equal and opposite. Therefore, equating the RHSs of equations (1) and (2) we have:

k x 2 x 10^-6 C / r² = k x 3 x 10^-6 C / (3-r)²

Cancelling out the common terms, we get:

2 / r² = 3 / (3-r)²

(3-r)² / r² = 3 / 2 = 1.5

(3-r)² / r² = 1.5

Taking the square root of both sides, we get:

(3-r) / r = 1.225

3 - r = r x 1.225

3 = r + (r x 1.225) = r x 2.225

3 = r x 2.225

r = 3 / 2.225 = 1.35

r = 1.35----------------------(answer)

Therefore, x coordinate of point P is given by:

x = -2 m + 1.35 m = -0.65 m

Therefore, coordinates of point P are (-0.65m, 0 m)--------------(answer)

Answers:

r = 1.35 m

Coordinates of P are (-0.65 m, 0 m)