Home
Class 12
PHYSICS
Calculate the minimum electrostatic forc...

Calculate the minimum electrostatic force between two charged particles separated at a distance of 1 cm in air.

Text Solution

Verified by Experts

As we know that the minimum charge is an electronic
Charge , i.e. ` " " e= 16 xx 10 ^(-19 ) C`
Given ` " " r =1 cm = 10 ^(-2) m `
So, minimum electrostatic force is the force between two electons or between an electron andn a proton
` because " " F = (kq_1q_2)/(r^(2)) rArr F = (ke^(2))/(r^(2))`
` therefore " " F = ((1.6 xx 10 ^(-19)^(2)))/((10^(-2) )^(2)) xx 9 xx 10 ^(9) `
` = (2.56 xx 9 xx 10 ^(9) xx 10 ^(-38))/(10^(-4))`
` F_(" min ") = 23.04 xx 10 ^(-25) N`
Promotional Banner

Similar Questions

Explore conceptually related problems

Force between two charges separated by a distance r varies as r^(2) ?

The ratio of gravitational force and electrostatic force between two electrons separated by a distance of 10 cm is of the order of 10 ^(42 )

State Coulomb's law of electric force between two charged bodies.

Gauss' law governs the force between two charges.

The force of attraction between two charges each of magnitude 3C separated by a distance 2m in air is

The electrostatic force between two charges Q_(1) and Q_(1) separated by a distance r is given by F="k"(Q_(1)Q_(2))/(r^(2)) . The constant k depends on