Question

Red blood cells often can be charged. Two red blood cells are separated by 1.46 m and have an attractive electrostatic force of 0.965 N between them. If one of the red blood cells has a charge of +8.32 10-6 C, what is the sign and magnitude of the second charge

Answer 1

`2.75\cdot 10^(-5)`, negative charge

Step-by-step explanation:

The magnitude of the electrostatic force between the two red blood cells is given by

`F=k (q_1 q_2)/(r^2)`

where

k is the Coulomb's constant

q1 and q2 are the charges of the two cells

r is their separation

In this problem we know that

`F=0.965 N`

r = 1.46 m

`q_1 = 8.32\cdot 10^(-6) C`

Solving for q2, we find

`q_2 = (Fr^2)/(kq_1)=((0.965)(1.46)^2)/((9\cdot 10^9)(8.32\cdot 10^(-6)))=2.75\cdot 10^(-5)`

And the force between the two cells is attractive, so since the first cell has positive charge, this second cell must have negative charge.