Question

Two tiny, spherical water drops, with identical charges of -7.08 × 10-16 C, have a center-to-center separation of 1.00 cm. (a) What is the magnitude of the electrostatic force acting between them? (b) How many excess electrons are on each drop, giving it its charge imbalance?

Answer 1

Answer:

(a) Electrostatic force F=4.51×10⁻¹⁷N

(b) Number of electron n=4425 electrons

Explanation:

Given data

Charges q₁=q₂= -7.08×10⁻¹⁶C

Distance r=1.00cm =0.01 m

To find

(a) Electrostatic force F

(b) Number of electron n

Solution

For (a) Electrostatic force

From Coulombs law we know that

$F=k\frac{q_{1}q_{2} }{r^{2} }\\ As\\ q_{1}=q_{2}=q\\So\\F=k\frac{q^{2}}{r^{2} }\\F=(8.99*10^{9})\frac{(-7.08*10)^{-16})^{2}}{(0.01)^{2} }\\F=4.51*10^{-17}N$

For (b) number of electron n

The number of electron on the drop that giving it its imbalance is the total charge divided by charge of electron

As we now that charge of electron e= -1.6×10⁻¹⁹C

$n=q/e\\n=\frac{-7.08*10^{-16} C}{-1.6*10^{-19} C}\\ n=4425electrons$