Question

If the third charge (–| q 3 |) is placed at point P, but not held fixed, it will experience a force and accelerate away from the other two charges. If q3 = q = –1.0 nC, the mass of the third charge m3 is 5.0 x 10-12 kg, and s = 4.0 cm, what will the speed of charged particle #3 [in m/s] be after it has moved a very large distance away?

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

A

The potential of this system is  $U=6.75*10^{-7}J$

B

The electric potential at point p is $V_p= -900V$

C

The work required is  $W= 9*10^{-7}J$

D

The speed of the charge is  $v=600m/s$

Explanation:

A sketch to explain the question is shown on the second uploaded image

Generally the potential energy for a system of two charges is mathematically represented as

            $U = \frac{kq_1 q_2}{d}$

where k is the electrostatic constant with a value of  $k = 9*10^9 N m^2 /C^2$

           q is the charge with a value of  $q = 1*10^{-9}C$

           d is the distance given as   $d =5m$

Now we are given that  $q_1 = q$ and  $q_2 = 3q$ and

Now substituting values

             $U = \frac{9*10^9 *1*10^{-9} * 3*10^{-9}}{5}$

                $U=6.75*10^{-7}J$

The electric potential at point P is mathematically obtained with the formula

             $V_p = V_{-q} + V_{-3q}$

I.e the potential at $q_1$ plus the potential at  $q_2$

Now potential at $q_1$ is mathematically represented as

                   $V_{-q} = \frac{-kq}{s}$

and the potential at $q_2$ is mathematically represented as

                        $V_{-3q} = \frac{-3kq}{s}$

Now substituting into formula for potential at  P

                  $V_p = \frac{-kq}{s} + \frac{-3kq}{s} = -\frac{4kq}{s}$

                       $= \frac{4*9*10^9 *1*10^{-9}}{4*10^{-2}}$

                      $V_p= -900V$

The Workdone to bring the third negative charge is mathematically evaluated as

                $W =\Delta U = \frac{kq_1q_3}{s} + \frac{kq_2q_3}{s}$

                                 $= \frac{kq*q}{s} + \frac{kq*3q}{s}$

                                $= \frac{4kq^2}{s}$

                               $= \frac{4* 9*10^9 * (1*10^{-9})^2}{4*10^{-2}}$

                              $W= 9*10^{-7}J$

From the Question are told that the charge $q_3$ would a force and an acceleration which implies that all its potential energy would be converted to kinetic energy.This can be mathematically  represented as

                   $\Delta U = W = \frac{1}{2} m_{q_3} v^2$

                         $9*10^{-7} = \frac{1}{2} m_{q_3} v^2$

Where $m_{q_3} = 5.0*10^{-12}kg$

Now making v the subject we have

                 $v = \sqrt{\frac{9*10^{-12}}{5*10^{-12}*0.5} }$

                     $v=600m/s$