Answer:
(a) 2.88×10¹¹ N/C.
(b) 1.442×10¹⁰ N/C
(c) 46.14×10⁻⁹ N
Step-by-step 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