Question

Two point charges, the first with a charge of +3.50 10-8 C and the second with a charge of -4.50x10-6C, are separated by 25.0 cm. Find
the magnitude of the electrostatic force experienced by the positive charge.

a. 6.32 N
b. 2.27 N
c. 3.42 N
d. 4.45 N

Answer 1

Answer:

b. 2.27 N

Explanation:

Fe = (k)(Q1)(Q2) / d²

k ≈ 9 × 10⁹ N m² / C² [Coulomb's Constant] or 8.98 × 10⁹ (nearest hundredth).

q = particle charge [Coulombs; C]

d = distance [Meters; m]

Fe = Electrostatic force [Newtons; N]

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Fe = (9×10⁹)(3.5 × 10^(-8)(-4.5 × 10^(-6))/(.25)² N =

(9)(3.5)(-4.5) × (10^(9))(10^(-8))(10^(-6) / (.25)² N =

(9)(3.5)(-4.5) × (10^(9 - 8 - 6)) / (.25)² N =

(-141.75) × 10^(-5) / (.0625) N =

141.75 × 10^(-5) / (.0625) N

[opposite direction] =

(1.4175 × 10^(2)) × 10^(-5) / (.0625) N

[opposite direction] =

1.4175 × 10^(-3) / .0625 N

[opposite direction] =

1.4175 × 10^(-3) / 6.25 × 10^(-2) N =

14.175 / 6.25 N [opposite direction] =

2.268 N ≈ 2.27 N [opposite direction].