A point charge of 5.9 muC is placed at the origin (x_1 = 0 of a coordinate syste

Question

A point charge of 5.9 muC is placed at the origin (x_1 = 0 of a coordinate system, and another charge of -19 muC is placed placed on the x-axis at x_2 = 0.21 m. Where on the x-axis can a third charge be placed in meters so that the net force on itis zero? X_3 = -0.134 There are three distinct regions for this problem on the x-axis - what are they? Which region must the third charge go in order to make sure the force can vanish? To start with, consider the negative x-axis; the magnitude of the force from the charge at the origin will always be larger than the charge on the other side. Will the force ever vanish on the negative x-axis? What if both charges are positive; that is, what if the second charge is 1.9 muC?

Explanation / Answer

Ans:-

You are asking where the field is zero. Since the charges are opposite in sign the zero point is not between them. It will be closer to the smaller charge.

Let x be the distance from this charge...so (0.21 + x) is the distance to the other chrage

Now using E = k*q/r^2 we have k*q1/x^2 = k*q2/(0.21+x)^2

Therefore q1*(0.21 + x)^2 = q2*x^2

Expanding and removing the C for simplicity we have 1.90*(x^2 + 0.42x + 0.0441) = 5.90x^2

5.9x^2 - 1.9x^2-0.798x – 0.0848 =0

4x^2 -0.798x -0.0848 = 0

X= 0.276m
So the charge must be at 0.276 +0.21 = 0.486m

B] Therefore q1*(0.21 + x)^2 = -q2*x^2

Expanding and removing the C for simplicity we have 1.90*(x^2 + 0.42x + 0.0441) =- 5.90x^2

5.9x^2 + 1.9x^2 + 0.798x + 0.0848 =0

7.8x^2 +0.798x + 0.0848 = 0

X=-0.0511538 -0.0908575i
there is imaginary value

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