Question

Discuss the symmetry between the laws of Charles Coulomb and Sir Isaac Newton regarding forces

Matt forgot to put the fabric softener in the wash. As his socks tumbled in the dryer, they became charged. If a small piece of lint with a charge of +1.25 E -19 C is attracted to the socks by a force of 3.0 E -9 N, what is the magnitude of the electric field at this location?

Suppose an electric field with a magnitude of 2.8 E 4 N/C at a certain location, points due south. If a charge of -4.0 E -6 C is brought to the location, what is the electrostatic force that acts on this c

Answer 1

Base on the question, the answer and the explanation would be, Newton's law of gravitation has force proportional to the product of the masses of the two bodies and inversely proportional to the square of the distance between them. Coulomb's law is similar, with the charges replacing the masses; however, you can have repulsion (+)(+) or (-)(-) with electrostatics but only attraction with gravitation.

Answer 2

Explanation :

(1) Charles Coulomb discovered the electrostatic force acting on two charged bodies.

According to him " the force of attraction or repulsion is directly proportional to the product of their charges and inversely proportional to the square of distance between them.''

$F=\dfrac{1}{4\pi \epsilon_0}\dfrac{q_1q_2}{r^2}$

where

$\dfrac{1}{4\pi \epsilon_0}$ is electrostatic constant.

While Sir Isaac Newton discovered the gravitational forces acting on two masses.

According to him " there exists a force in the universe which attracts every other object with a force which is equal to the product of their masses and inversely proportional to the square of the distance between them."

$F=G\dfrac{m_1m_2}{r^2}$

where,

G is universal gravitational constant.

(2) It is given that,

Charge, $q=1.25\times 10^{-19}\ N$

Force, $F=3\times 10^{-9}\ N$

We know that the relation between electric field and electric force is F = q E.

So, $E=\dfrac{F}{q}$

$E=\dfrac{3\times 10^{-9}\ N}{1.25\times 10^{-19}\ C}}$

$E=2.4\times 10^{10}\ N/C$

(3) It is given that,

Electric field, $E=2.8\times 10^4\ N/C$

Charge, $q=-4\times 10^{-6}\ C$

Since, F = q E

So, $F=-4\times 10^{-6}\ C\times 2.8\times 10^4\ N/C$

$F=-11.2\times 10^{-2}\ N$

Negative sign shows the force is attractive.

Hence, this is the required solution.