Question

At locations A and B, the electric potential has the values = 1.19 V and VB = 5.45 V, respectively. A proton released from rest at one of these locations passes through the other location From which location was the proton released? O A What is its speed when it passes through the other location? speed: m/s

Answer 1

Let A and B be two points with potentials 1.19v and 5.45v keep. So, B is at a higher potential than A. we know an electron moves from low potential to high potential and a proton (or positive charge) moves from high potential to low potential. (i) So from thus we can say that the proton was released from the higher potential i. e. , B (ii)From Newton's 2nd law and laws of electrostatics we know the difference in K. E. between these two points are 1/2 m_p V_A^2 - 1/2 m_p V_B^2 = q_e V---(1) where m_p is the mass of proton = 1.67 times 10^-27 kg V_A = velocity of the particle at A V_B = vel. Of the particle at B = 0 q_e = charge of proton = 1.6 times 10^-19 C V = voltage difference between A and B \r\n Substituting all values in (1) 1/2 times 1.67 times 10^-27 times V_A^2 = 1.6 times 10^-19 times (5.45 - 1.19)