Answer:
20.353125 V
Explanation:
m = Mass of proton = [tex]1.67\times 10^{-27}\ kg[/tex]
q = Charge of proton = [tex]1.6\times 10^{-19}\ C[/tex]
[tex]v_A[/tex] = Velocity of proton at A = 50 km/s
[tex]v_B[/tex] = Velocity of proton at B = 80 km/s
We have the relation by balancing the energy
[tex]\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[/tex]
The potential difference is 20.353125 V
The potential difference is 20.35 V.
The potential difference is determined by applying the principle of conservation of energy as shown below;
ΔK.E = ΔU
¹/₂m(Vb² - Va²) = q(Vb - Va)
Where;
m is the mass of the proton
q is the charge
Va is velocity of the proton at A
Vb is velocity of the proton at B
[tex]V_b - V_a= \frac{m}{2q} (V_b^2 - V_a^2)\\\\V_b - V_a= \frac{1.67 \times 10^{-27}}{2 \times 1.6 \times 10^{-19}} (80000^2 - 50000^2)\\\\V_b - V_a= 20.35 \ V[/tex]
Thus, the potential difference is 20.35 V.
Learn more about potential difference here: https://brainly.com/question/3406867
