An electron moving along the x axis has an initial speed of 1 × 106 m/s at the origin. Its speed is reduced to 5 × 105 m/s at the point xP , 3 cm away from the origin. Calculate the magnitude of the potential difference between this point and the origin. The mass of the electron is 9.109 × 10−31 kg and the elemental charge is 1.602 × 10−19 C. Answer in units of V.

By conservation of energy principle, we know that the potential difference between two points is equal to the change in the kinetic energy between these points.

[tex]-\Delta U = \Delta KE\\\\-e\Delta V = \frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2[/tex]

Hence,

[tex]1.602*10^{-19}\Delta V = \frac{1}{2}*9.11*10^{-31}((5*10^5)^2 - (1*10^6)^2)[/tex]

⇒ [tex]\Delta V = -2.13[/tex] V

mickymike92 mickymike92 · Answer 2 · 2022-02-22T14:06:37+01:00

Based on the data given, the magnitude of the potential difference between this point and the origin is -2.13 V.

What is the electric potential?

Electric potential is the work done in bringing a unit positive charge from infinity to that point against the action of the field.

Electrical energy is related to potential difference as shown in the formula below:

E= qV

From the law of conservation of energy, the potential difference between the two points is equal to the change in the kinetic energy of the electron.

Therefore;

qV = ΔKE

ΔKE = 1/2mv² - 1/2mu²

q = 1.602 × 10⁻¹⁹ C

v = 5 × 10⁵ m/s

u = 1 * 10⁶ m/s

m = 9.109 * 10⁻³¹ kg

V = ΔKE/q

V = {1/2 * 9.109 * 10⁻³¹ kg [(5 × 10⁵)² - (1 * 10⁶)²]}/ 1.602 × 10⁻¹⁹ C

V = -2.13 V

Therefore, the magnitude of the potential difference between this point and the origin is -2.13 V.

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