Two cahrged particles ar eplaced at a distance 1.0 cm apart. What is the minimum possible magnitude of the electric force acting on eachcharge?

Open in App

Solution

Minimum charge on body is equal to the cahrge of electron $= 1.6 \times 10^{- 19} C,$
$r = 1.0 c m = 10^{- 2} m F = \frac{9 \times 10^{9} \times (1.6 \times 10^{19})^{2}}{(10^{- 2})^{2}} = 2.3 \times 10^{- 24} N .$

Suggest Corrections

Similar questions

Q. Two charged particles are placed at a distance

1.0 c m

apart. What is the minimum possible magnitude of the electric force acting on each charge?

Q. Two charged particles are placed 1.0 cm apart. What is the minimum possible magnitude of the electric force acting on each charge?

Two charged particles are separated at a distance of 1 cm.what is the minimum possible magnitude of the electric force acting on each particle ?

Q. Two charged particles are placed at a distance of 1.0 cm apart. What is the minimum possible magnitude of electrostatic force acting on each charge

Two equal charges ar eplaced at a separation of 1.0 m. What should be the magnitude of the charges so that the force between them equals the weight of a 50 kg person?

Join BYJU'S Learning Program

Two cahrged particles ar eplaced at a distance 1.0 cm apart. What is the minimum possible magnitude of the electric force acting on eachcharge?

Minimum charge on body is equal to the cahrge of electron =1.6×10−19C, r=1.0cm=10−2mF=9×109×(1.6×1019)2(10−2)2=2.3×10−24N.

Minimum charge on body is equal to the cahrge of electron $= 1.6 \times 10^{- 19} C,$
$r = 1.0 c m = 10^{- 2} m F = \frac{9 \times 10^{9} \times (1.6 \times 10^{19})^{2}}{(10^{- 2})^{2}} = 2.3 \times 10^{- 24} N .$