Question

Consider an electron that is 10^-10 m from an alpha particle (q = +3.2 times 10^-19 C). (a) What is the electric field due to the alpha particle at the location of the electron? (b) What is the electric field due to the electron at the location of the alpha particle? (c) What is the electric force on the alpha particle? on the electron?

Answer 1

Answer:

(a) 2.88×10¹¹ N/C.

(b) 1.442×10¹⁰ N/C

(c) 46.14×10⁻⁹ N

Explanation:

(a) Electric field due to alpha particle at the location of the electron

E = Kq/r²....................... Equation 1

Where E = Electric field, q = charge of the alpha particle, K = proportionality constant, r = distance between the electron and the alpha particle.

Given: q = 3.2×10⁻¹⁹ C, r = 10⁻¹⁰ m.

Constant: K = 9×10⁹ Nm²/C²

Substituting into equation 1

E = 9×10⁹(3.2×10⁻¹⁹)/(10⁻¹⁰)²

E = 28.8×10⁻¹⁰/10⁻²⁰

E = 2.88×10¹¹ N/C.

(b) Electric Field due to the electron at the location of the alpha particle

E₁ = Kq₁/r²............................... Equation 2

Where,

E₁ = Electric Field due to the electron at the location of the alpha particle.

q₁ = charge of an electron.

Given: r = 10⁻¹⁰ m

Constant: K = 9×10⁹ Nm²/C², q₁ = 1.602×10⁻¹⁹ C

Substituting into equation 2

E₁ = 9×10⁹×1.602×10⁻¹⁹/(10⁻¹⁰ )²

E₁ = 14.42×10⁻¹⁰/10⁻²⁰

E₁ = 14.42×10¹⁰

E₁ = 1.442×10¹⁰ N/C.

(c) The electric force on the alpha particle and on the electron.

F = Kqq₁/r² ................................ Equation 3

Given: q = 3.2×10⁻¹⁹ C, r = 10⁻¹⁰ m,

Constant:  K = 9×10⁹ Nm²/C², q₁ = 1.602×10⁻¹⁹ C

Substituting these values into equation 3

F =9×10⁹( 3.2×10⁻¹⁹ )(1.602×10⁻¹⁹)/(10⁻¹⁰)²

F = 46.14×10⁻²⁹/10⁻²⁰

F = 46.14×10⁻⁹ N

Thus the electric force = 46.14×10⁻⁹ N