Answer:
The net force on the electron is given as:
F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i
Explanation:
Given:
charge on rod along x-axis = Q₁ = -15 x 10⁻⁶ C
charge on rod along y-axis = Q₂ = 15 x 10⁻⁶ C
distance of electron from rod 1 = r₁ = 0.4 m
distance of electron from rod 1 = r₂ = 0.4 m
charge on electron = q = -1.6 x 10⁻¹⁹ C
ε° = 8.85 x 10⁻¹² C²/Nm²
Electric force on charge due to rod 1:
F₁ = qE = 1/4πε°(qQ₁/r₁²)
F₁ = (9 x 10⁹ x -1.6 x 10⁻¹⁹ x -15 x 10⁻⁶)/0.4²
F₁ = 1.35 x 10⁻¹³ N
Negative negative repels each other so the rod will Force the electron in positive y-direction.
F₁ = 1.35 x 10⁻¹³ N j
Electric force on charge due to rod 2:
F₂ = qE = 1/4πε°(qQ₂/r₂²)
F₂ = (9 x 10⁹ x -1.6 x 10⁻¹⁹ x 15 x 10⁻⁶)/0.4²
F₂ = - 1.35 x 10⁻¹³ N
Opposite charges attract each other so the rod will force the electron in negative x-direction.
F₂ = - 1.35 x 10⁻¹³ N i
Net Force:
F = F₁ + F₂
F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i
