Starting from rest, a proton was accelerated by an electric field. After a distance of 1.35 m the proton reaches a speed of 1.97 ✕ 105 m/s (mp = 1.67 ✕ 10−27 kg, qp= 1.60 ✕ 10−19 C). a) What is the final kinetic energy of the proton? b) What is the potential energy difference of the proton? c) What is the magnitude of the electric field?

A proton (m=1.67×10-27kg, q=+1.60×10-19C) is moving towards a stationary point charge (Q = +1.50 μC). The proton is initially has a speed of vi=2.50×106m/s, and is a distance of 6.50 meters away from the point charge.(a)How fast is the proton moving (in m/s) when it is 3.50 meters away from thepoint charge? Use conservation of energy.(b)How far is the proton from the point charge (in cm) when it comes to a stop?

A From Equation 25.6, we have the following. AV = -Ed = -(7.2 × 104 v/m)(0.60 m) = B V From Equation 25.3, we have the following. AU = qAV = eAV = (1.6 x 10-19 C)(AV) | The negative sign means that the potential energy of the system decreases as the proton moves in the direction of the electric field. This decrease is consistent with the conservation of energy in an isolated system: as the proton accelerates in the direction of the field, it gains kinetic energy and at the same time the system loses electric potential energy. The increase in kinetic energy of a charged particle in an electric field is exploited in many devices, including electron guns for TV picture tubes and particle accelerators for research in particle physics. You can predict and observe the speed of the proton as it arrives at the negative plate for random values of the electric field using this Interactive Example. Hints: Getting Started | I'm Stuck Exercise 25.2 For the previous example, apply the principle of…

A proton is fired from very far away directly at a fixed particle with charge q = 1.12 x 10-18 C. If the initial speed of the proton is 3.0 x 105 m/s, what is its distance of closest approach to the fixed particle? The mass of a proton is -27 kg. m 1.67 X 10