IP The point charges in the figure below have the following values: q 1 = 2.1 C,

Question

IP The point charges in the figure below have the following values: q1 = 2.1 C, q2 = 6.2 C, q3 = -0.87 C.

Assume that charge q1 is at the origin and q3 lies on the positive x-axis.

PART A: Given that the distance d in the figure is 4.30 cm, find the magnitude of the net electrostatic force exerted on the point charge q1.

Express your answer using two significant figures.

Part B

Find the direction of the net electrostatic force exerted on the point charge q1 in Part A.

Express your answer using three significant figures.

Part C

How would your answer to Part A change if the distance d were doubled?

A) increased by the factor of 2 B) increased by the factor of 4 C) decreased by the factor of 2 D) decreased by the factor of 4 E) stay the same 4 13 91

Explanation / Answer

Part A:

Given that -
q1 = 2.1 uC
q2 = 6.2 uC
q3 = -0.87 uC

Force, F between the two charges is -

F=k*q1*q2/r^2

where, k = 9*10^-3
vector Force on q1 from q2: "third quadrant"

F=9*10^-3*(2.1)*(6.2)/ (4.3/100)^2 = 63.37 N

Componets of F: acts @ 240 degrees from x-axis
Fi = 63.37* Cos(240) = - 31.69i N <---
Fj= 63.37* Sin(240)= - 54.88j N

Force on q1 from q3
Fi = 9*10^-3*(2.1)*(0.87)/ (4.3/100)^2 = 8.89 N --->

Sum x = -31.69 i + 8.89 i = -22.8 i N
Sum y = -54.88 j N
therefore -
x - component = -22.8
y - component = -54.88

So, the resultant force on q1
= sqr(x^2+y^2) = 59.43 N

Part B:

angle with positive x-axis is:
tan^-1 (y/x) = tan^-1 (-54.88 / -22.8) = 292 deg.

Part C:

Force is inversely proportional to the sqare of distance between the charges.

So, when distance is doubled, the force will decrease by a factor '4'.

So, option (D) is your correct answer.

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