Complete Question
The complete question is shown on the first uploaded image
Answer:
A
The potential of this system is [tex]U=6.75*10^{-7}J[/tex]
B
The electric potential at point p is [tex]V_p= -900V[/tex]
C
The work required is [tex]W= 9*10^{-7}J[/tex]
D
The speed of the charge is [tex]v=600m/s[/tex]
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
[tex]U = \frac{kq_1 q_2}{d}[/tex]
where k is the electrostatic constant with a value of [tex]k = 9*10^9 N m^2 /C^2[/tex]
q is the charge with a value of [tex]q = 1*10^{-9}C[/tex]
d is the distance given as [tex]d =5m[/tex]
Now we are given that [tex]q_1 = q[/tex] and [tex]q_2 = 3q[/tex] and
Now substituting values
[tex]U = \frac{9*10^9 *1*10^{-9} * 3*10^{-9}}{5}[/tex]
[tex]U=6.75*10^{-7}J[/tex]
The electric potential at point P is mathematically obtained with the formula
[tex]V_p = V_{-q} + V_{-3q}[/tex]
I.e the potential at [tex]q_1[/tex] plus the potential at [tex]q_2[/tex]
Now potential at [tex]q_1[/tex] is mathematically represented as
[tex]V_{-q} = \frac{-kq}{s}[/tex]
and the potential at [tex]q_2[/tex] is mathematically represented as
[tex]V_{-3q} = \frac{-3kq}{s}[/tex]
Now substituting into formula for potential at P
[tex]V_p = \frac{-kq}{s} + \frac{-3kq}{s} = -\frac{4kq}{s}[/tex]
[tex]= \frac{4*9*10^9 *1*10^{-9}}{4*10^{-2}}[/tex]
[tex]V_p= -900V[/tex]
The Workdone to bring the third negative charge is mathematically evaluated as
[tex]W =\Delta U = \frac{kq_1q_3}{s} + \frac{kq_2q_3}{s}[/tex]
[tex]= \frac{kq*q}{s} + \frac{kq*3q}{s}[/tex]
[tex]= \frac{4kq^2}{s}[/tex]
[tex]= \frac{4* 9*10^9 * (1*10^{-9})^2}{4*10^{-2}}[/tex]
[tex]W= 9*10^{-7}J[/tex]
From the Question are told that the charge [tex]q_3[/tex] 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
[tex]\Delta U = W = \frac{1}{2} m_{q_3} v^2[/tex]
[tex]9*10^{-7} = \frac{1}{2} m_{q_3} v^2[/tex]
Where [tex]m_{q_3} = 5.0*10^{-12}kg[/tex]
Now making v the subject we have
[tex]v = \sqrt{\frac{9*10^{-12}}{5*10^{-12}*0.5} }[/tex]
[tex]v=600m/s[/tex]
