Answer:
[tex]V_B-V_A=-20736-0=-20736volt[/tex]
Explanation:
We have given charge on the particle [tex]q=-4\mu C=-4\times 10^{-6}C[/tex]
Mass of the charge particle [tex]m=3.2\times 10^{-6}kg[/tex]
From energy of conservation kinetic energy will be equal to potential energy
So at point A
[tex]\frac{1}{2}mv^2=qV[/tex]
At point a velocity is zero
So [tex]\frac{1}{2}(3.2\times10^{-6} )0^2=-4\times 10^{-6}V_a[/tex]
[tex]V_A=0volt[/tex]
At point B velocity will be 72 m/sec
So [tex]\frac{1}{2}\times 3.2\times 10^{-6}72^2=-4\times 10^{-6}V_b[/tex]
[tex]V_B=-20736volt[/tex]
So [tex]V_B-V_A=-20736-0=-20736volt[/tex]
