Answer:
[tex]V_B-V_A=+1631.47 V[/tex]
Explanation:
We are given that
[tex]q=-1.7\mu C=-1.7\times 10^{-6} C[/tex]
Mass,m=[tex]3.0\times 10^{-6}kg[/tex]
Speed,v=43 m/s
We have to find the potential difference between A and B.
By law of conservation of energy
Kinetic energy=Potential energy
[tex]\frac{1}{2}mv^2=q(V_B-V_A)[/tex]
Substitute the values
[tex]\frac{1}{2}(3\times 10^{-6}(43)^2=1.7\times 10^{-6}(V_B-V_A)[/tex]
[tex]V_B-V_A=\frac{3\times 10^{-6}}{2\times 1.7\times 10^{-6}}(43)^2[/tex]
[tex]V_B-V_A=1631.47 V[/tex]
Negative charge is always accelerated towards high potential.Therefore,potential at point B is higher than the potential at point A.
Hence,[tex]V_B-V_A=+1631.47 V[/tex]
