Answer: 247.67 V
Explanation:
Given
Potential At A [tex]V_a=382\ V[/tex]
Potential at [tex]V_c=785\ V[/tex]
when particle starts from A it reaches with velocity [tex]v_b[/tex] at Point while when it starts from C it reaches at point B with velocity [tex]2v_b[/tex]
Suppose m is the mass of Particle
Change in Kinetic Energy of particle moving under the Potential From A to B
[tex]q\cdot \left ( V_a-V_b\right )=0.5m\cdot (v_b)^2----1[/tex]
Change in Kinetic Energy of particle moving under the Potential From C to B
[tex]q\cdot \left ( V_c-V_b\right )=0.5m\cdot (2v_b)^2-----2[/tex]
Divide 1 and 2 we get
[tex]\frac{V_a-V_b}{V_c-V_b}=\frac{v_b^2}{4v_b^2}[/tex]
on solving we get
[tex]V_b=\frac{4}{3}\cdot V_a-\frac{1}{3}\cdot V_c[/tex]
[tex]V_b=\frac{743}{3}=247.67\ V[/tex]
