A point charge q 1 = 3.90 nC is placed at the origin, and a second point charge

Question

A point charge q1 = 3.90 nC is placed at the origin, and a second point charge q2 = -3.05 nC is placed on the x-axis at x=+ 20.5 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.)

Part A

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

3.1761•107

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Part B

Where should q3 be placed between q1 and q2 to make the potential energy of the system equal to zero?

A point charge q1 = 3.90 nC is placed at the origin, and a second point charge q2 = -3.05 nC is placed on the x-axis at x=+ 20.5 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.)

Part A

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

3.1761•107

SubmitMy AnswersGive Up

Incorrect; Try Again; 2 attempts remaining

Part B

Where should q3 be placed between q1 and q2 to make the potential energy of the system equal to zero?

Explanation / Answer

1) The potential energy of the system is given by the formula U= kQq/r ^2where k is coloumb constant . However while finding the potential energy we may calculate this way as well: U= kQ1Q3/r+ kQ2 Q3/r taking all three charges in a same line (considering the signs as it chAnges the calculation)

=k(Q1Q3+Q2Q3/r)

=9×10^9( 3.90*10^-9 *2.10×10-9/10.5 +2.10*10-9*3.05*10-9/10)

=9*10-9 (0.78-0.64) 10^-18

U= 0.147×10^-9 J

b) The potential energy will only become zero when q3 is placed at 11.58 cm away from q1 on x axis. It can be solved as follows

Potential i.e V= kQ/r

=3.90×10^-9×k/x + (-3.05)×10^-9×k/20.5-x (assuming yhe distance between q1 and q3 to be x such that the remaining distance between q3 and q2 becomes 20.5-x)

As we take V=0 because that is how we will get to knos where the potential becomes zero.. The position can be found out.

3.90×10^-9×k/x =3.05×10^-9×k/20.5-x

Which equals to after solving both sides

1.3(20.5-x)=x

26.65=2.3x so x =11.58 cm

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