Answer:
a)2.59 ×10⁵ N/C
Direction is to the right.
b) 4.14 × 10⁻¹⁴ N , direction is to the Left.
Explanation:
Given:
Charge [tex]q_{1}[/tex] = [tex]1.6\times 10^{-7}C[/tex]
Charge [tex]q_{2}[/tex] = [tex]-6.8\times 10^{-8}C[/tex]
Distance from the negative charge = 0.05 m
Distance to [tex]q_{1}[/tex] = 0.032 m
a) Electric field = E = [tex]k[\frac{q_{1}}{r_{1}^{2}} + \frac{q_{2}}{r_{2}^{2}}][/tex]
k = Coulomb's constant = 9 × 10⁹ N m²/C²
E = [tex](9\times 10^{9})[\frac{1.6 \times 10^{-7}}{(0.032)^{2}} + \frac{6.8 \times 10^{-8}}{(0.05)^{2}}][/tex] = (1.40625 × 10⁴ + 24.48 × 10⁴ ]
= 2.59 ×10⁵ N/C
Direction is to the right.
b) Force on the electron = Charge on the electron × Electric field E.
⇒ F = ( 1.6 × 10⁻19)(2.59 ×10⁵) = 4.14 × 10⁻¹⁴ N , direction is to the Left.
