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

Answer:

$2.75\cdot 10^{-5}$, negative charge

Explanation:

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

$F=k \frac{q_1 q_2}{r^2}$

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

$q_2 = \frac{Fr^2}{kq_1}=\frac{(0.965)(1.46)^2}{(9\cdot 10^9)(8.32\cdot 10^{-6})}=2.75\cdot 10^{-5}$

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