Question

ce only. At point A the proton moves with a speed of 50 km/s. At point B the speed of the proton is 80 km/s. Determine the potential difference .A proton (mass = 1.67 10–27 kg, charge = 1.60 10–19 C) moves from point A to point B under the influence of an electrostatic force only. At point A the proton moves with a speed of 50 km/s. At point B the speed of the proton is 80 km/s. Determine the potential difference .

Answer 1

Answer:

20.353125 V

Explanation:

m = Mass of proton = $1.67\times 10^{-27}\ kg$

q = Charge of proton = $1.6\times 10^{-19}\ C$

$v_A$ = Velocity of proton at A = 50 km/s

$v_B$ = Velocity of proton at B = 80 km/s

We have the relation by balancing the energy

$\dfrac{1}{2}m(v_B^2-v_A^2)=q(V_B-V_A)\\\Rightarrow V_B-V_A=\dfrac{1}{2q}m(v_B^2-v_A^2)\\\Rightarrow V_B-V_A=\dfrac{1}{2\times 1.6\times 10^{-19}}\times 1.67\times 10^{-27}(80000^2-50000^2)\\\Rightarrow V_B-V_A=20.353125\ V$

The potential difference is 20.353125 V