Question

An attractive force of |1N| exists between two charged particles. The magnitude of both charges is the same. If the particles are 0.25m apart what is the value and sign of the charge for both particles.

A. (- 2.6x10-6 C) and (+ 2.6x10-6 C)

B. (+ 2.7x10-11 C) and (- 2.7x10-11 C)

C. (- 5.27x10-6 C) and (+ 5.27x10-6 C)

D. (- 6.94x10-12 C) and (- 6.94x10-12 C)

Answer 1

A: -2.6 × 10^(-6) C and 2.6 × 10^(-6) C

Step-by-step explanation:

According to coulombs law, the formula for the force between 2 charged particles is;

F = kq1•q2/r²

Where;

F is Force = |1N|

q1 and q2 are the charges of both particles

k is coulombs constant = 9 × 10^(9) N.m²/C²

r is distance between particles = 0.25

Now, the force |1N| means -1 or +1

Thus;

-1 = 9 × 10^(9) × q1•q2/(0.25)²

Or

1 = 9 × 10^(9) × q1•q2/(0.25)²

Thus;

q1•q2 = -6.94 × 10^(-12)

Or q1•q2 = 6.94 × 10^(-12)

Now, we are told that the magnitude of both charges are the same.

Thus;

q² = 6.94 × 10^(-12)

q = √(6.94 × 10^(-12))

q = 2.6 × 10^(-6) C

Since from earlier, we saw that;

q1•q2 = -6.94 × 10^(-12)

Or q1•q2 = 6.94 × 10^(-12)

Thus means that one of the charges will be negative while the other will be positive.

Thus, the charges are;

-2.6 × 10^(-6) C and 2.6 × 10^(-6) C