Question

Figure (a) shows charged particles 1 and 2 that are fixed in place on an x axis. Particle 1 has a charge with a magnitude of lq1l-16e. Particle 3 of charge q3 16e is initially on the x axis near particle 2.Then particle 3 is gradually moved in the positive direction of the x axis. As a result, the magnitude of the net electrostatic force on particle 2 due to particles 1 and 3 changes. Figure (b) gives thex component of that net force as a function of the position x of particle 3. The scale of the x axis is set by xs 1.40 m. The plot has an asymptote of F2net-2.863 x 10- N as x → oo. As a multiple of e and including the sign, what is the charge q2 of particle 2? 25 x (m) 0 Number Units

Answer 1

Here we have to find both the charge and the magnitude of the charge q2.

First, we will decide the sign of the charge of q1 and q2.

Then we will try to find out the distance between the charge q1 and q2.

From the asymptote condition, we will get the magnitude of charge q2.

The law and principle used here:

• coulombs law:

  $\vec F= \frac{1}{4\pi \epsilon_o}\times \frac{q_1.q_2}{r^2}$

q1 and q2 are the charges respectively and r is the distance between the two particles.

• Like charges repel each other and unlike charges attract each other.
• The magnitude of e is:  $1.6\times 10^{-19}C$.

In the case of the position of q3 going to infinity, its effect will NOT be felt by q2. Given that the force in the positive direction, it is clear that the charge q1 is repelling it., Hence the charge of q1 and q2 must be same.

From the graph, it is seen that at a distance of 0.7m the net force experienced by the q2 is 0. In this case, also the charge q1 will be repelling the q2 as in the above case, the force on q2 due to q1 is in the positive x-direction. Moreover, the net force on q2 is 0, thus q3 must apply on it a force in the -ve x-direction, thus repelling it. Hence there is repulsion between the charges q2 and q3 also. Therefore q2 and q3 are also like charges. Given charge q3 is POSITIVE.

Hence we conclude that the charge q2 and q1 are also positive.

Let the distance between q1 and q2 = r1.

In the case of q3 at infinity, the net force on q2 is due to q1 only and is given by

$\\ \vec F= \frac{1}{4\pi \epsilon_o}\times \frac{16e.q_2}{r_1^2}= 2.863\times 10^{-25}N \\ \\ \\ (9 \times10^9 N. m^2/ C^2)\times( 16\times1.6\times 10^{-19}C)\times \frac{q_2}{r_1^2} =2.863\times 10^{-25}N \\ 2.304\times 10^{-8}Nm^2/C\times \frac{q_2}{r_1^2}=2.863\times 10^{-25}N\\ \\ \\\frac{q_2}{r_1^2}=1.243\times 10^{-17}C/m^2$ ...............................Equation (a).

Now from the graph, when the distance between q2 and q3 =0.7m the net force on the charge q2 is ZERO.

$\\ \vec F_{21}= \frac{1}{4\pi \epsilon_o}\times \frac{16e.q_2}{r_1^2}\\ \\ \\ \vec F_{23}= \frac{1}{4\pi \epsilon_o}\times \frac{16e.q_2}{(0.7m)^2} \\ \\$

These two force are equal and opposite hence

$\\ \frac{1}{4\pi \epsilon_o}\times \frac{16e.q_2}{r_1^2}= \frac{1}{4\pi \epsilon_o}\times \frac{16e.q_2}{(0.7m)^2} \\ \\ \frac{1}{r_1^2}=\frac{1}{(0.7m)^2}\\ \\ r_1=0.7 m$

putting this value of r1 in equation (a)

that is :

$\\\frac{q_2}{r_1^2}=1.243\times 10^{-17}C/m^2$

$\\\frac{q_2}{(0.7m)^2}=1.243\times 10^{-17}C/m^2$

$\\{q_2}=1.243\times 10^{-17}C/m^2\times (0.7m)^2\\ \\ q_2=6.088\times 10^{-18}C$

expressing in terms of e

$q_2=\frac{6.088\times 10^{-18}C}{1.6\times 10^{-19}C}e=38.05e$

charge o q2 is $6.088\times 10^{-18}C$ .

q2 is approximately 38e and positively charged.