1. Two equal point charges of +2.85×10−6 are placed 0.215 m a part (Part A) What...

Question

Question

1. Two equal point charges of +2.85×10⁻⁶ are placed 0.215 m a part

(Part A) What are the magnitude of the force each charge exerts on the other?

2. Two small spheres spaced 50.0 cm apart have equal charge.

(Part A) How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 5.21×10⁻²¹ N ?

Answer 1

Answer #1

1)

Force, F = (9 x 10^9)(2.85 x 10^-6)^2/0.215^2 = 1.58 N

2)

Force, F = k q^2/r^2

5.21 x 10^-21 = (9 x 10^9) q^2/0.5^2

Charge, q = 3.8 x 10^-16

Charge, q = ne

No. of electrons, n = (3.8 x 10^-16)/(1.6 x 10^-19) = 2377.65

Comment in case any doubt please rate my answer ...

Answer 2

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