Question

A point charge q1 = 4.10 nC is placed at the origin, and a second point charge q2 = -3.10 nC is placed on the x-axis at x=+ 21.0 cm . A third point charge q3 = 1.90 nC is to be placed on the x-axis between q1 and q2. (Take as zero the potential energy of the three charges when they are infinitely far apart.)

**Part A**

What is the potential energy of the system of the three charges
if *q*3 is placed at *x*=+ 10.5 cm ?

**Part B**

Where should *q*3 be placed between *q*1 and
*q*2 to make the potential energy of the system equal to
zero?

Answer #1

A) if q3 is placed at x=+ 10.5 cm

the distance between q1 and q2 is 21 cm

between q1 and q3 is 10.5 cm

between q2 and q3 is 10.5 cm

then the potential energy of the system of the three charges if q3 is placed at x=+ 10.5 cm is

B) Let q3 be placed at + x between q1 and q2

Then if the potential energy of the system is zero, then

Solving this quadratic equation gives 2 solutions

we have chosen x to be in between 0 and 0.21 m

Hence

is the correct solution

Therefore q3 has to be placed at x = + 7.48 cm to make the potential energy of the system equal to zero

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20.0 cm . A third point charge q3 = 2.10 nC is to be placed on the
x-axis between q1 and q2. (Take as zero the potential energy of the
three charges when they are infinitely far apart.)
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