Question

coulomb's law Ka dereviation​

Answer 1

Explanation:

We can show it with the following explanation. Let’s say that there are two charges q1 and q2. The distance between the charges is ‘r’, and the force of attraction/repulsion between them is ‘F’. Then

F ∝ q1q2

Or, F ∝ 1/r2

F = k q1q2/ r2

where k is proportionality constant and equals to 1/4 π ε0. Here, ε0 is the epsilon naught and it signifies permittivity of a vacuum. The value of k comes 9 × 109 Nm2/ C2 when we take the S.I unit of value of ε0 is 8.854 × 10-12 C2 N-1 m-2.

According to this theory, like charges repel each other and unlike charges attract each other. This means charges of same sign will push each other with repulsive forces while charges with opposite signs will pull each other with attractive force.

Answer 2

Answer: let Force=F.

Let the charges be q1 and q2.

Let the constant be= k.

So, we know that force is dependent on charges ka magnitude.

Therefore, F is directly proportional to q1×q2

Also, force is dependent on the distance between the 2 charges. So F is inversely proportional to r^2(as force between particles decreases on increasing distance)

So, overall: F is directly proportional to (q1×q2)/r^2

And to remove the "directly proportional" sign, we use constant k.

So, F= k×(q1×q2)/r^2. Where k=1/4π€•

Hope it helps!!!!!

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