Question

ASK YOUR TEACHER Red blood cells often can be charged. Two red blood cells are separated by 1.11 m and have an attractive electrostatic force of 0.985 N between them. If one of the red blood cells has a charge of 8.40 10-6 C, what is the sign and magnitude of the second charge, Q

Answer 1

Answer:

This can be explained using Coulomb's law.

F= k|Q|q ÷ r²

where F = 0.985N, r = 1.11, k = 8.99 × 10⁹, q = 8.40 × 10⁻⁶, Q = x

substituting the values into the equation above, we have

Q = (0.985) × (1.11)² ÷ (8.99 × 10⁹) × (8.40 × 10⁻⁶)

Q = 1.2136185 ÷ 75516

Q = 0.00001607101

Q = 1.61 × 10⁻⁵C

Since the force is an attractive force, the value will be negative,

∴ Q = -1.61 × 10⁻⁵C

Answer 2

Answer:

q2 = -1.61*10^-5 C.

Explanation:

It was given that,

F = 0.985N

q1 = +8.40 X10-6 C

q2 = ?

r = 1.11 m

k = 9 x 10^9 (standard)

It generally follows that, if force is attractive, charge will be negative.

force, F = kq1q2/r^2

0.985 = 9*10^9*8.40*10^-6*q2/1.11^2

75600q2 = 0.985*1.11^2

75600q2 = 1.2136

q2 = 1.2136/75600 = 1.60529

q2 = -1.61*10^-5 C.