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

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.

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

gets less

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.