Question

when charge 2 is 3.0 m away from charge 1, the strength of the electric force on charge 2 by charge 1 is 0.80 n. if instead, charge 2 were 6.0 m away from charge 1, what would be the strength of the electric force on charge 2 by charge 1 in that case?

Answer 1

The strength of the electrostatic force is inversely proportional to the square of the distance

between the charges. If the distance is doubled, the force is 1/4.

The new force is 0.80/4 = 0.20 N.

What is electrostatic force?

Electrostatic force is the attractive or repulsive force that exists between two charged particles. Also known as Coulomb interaction or Coulomb force. For example, the electrostatic forces between the protons and electrons of an atom are responsible for the stability of the atom.

The force acting along a line joining two point charges is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them.

$F=K |\frac{q_{1} q_{2}}{r^{2} } |$

In the above formula, k is arbitrary and can be chosen to be any positive value. Since k is a constant, I chose to give the value of k as follows:

Therefore, with q₁ and q₂ values ​​of 1 and r = 1 (two charges with 1 Coulomb charge each at a distance of 1 m), we get F = $9 \times 10^9$ N. In the above equation, ε₀ is given as the permittivity of free space and its value in SI units is $8.854\times10^{-12} C^{2} N^{-1} m^{-2}$.

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