Question

Two helium nuclei are separated by a distance 6 nm in
vacuum. What is the electrostatic repulsion between them?
(A) 25.6 PN
(B) 25.6 nN
(C) 3.6 PN
(D) 3.6 nN​

Answer 1

Answer: (A) 25.6pN

Explanation:

F = 9x10^9 (2x1.6x10^-19)^2 / (6x10^-9)^2

=> 25.6pN

Answer 2

Concept

A nucleus is the membrane- enclosed organelle within a cell that contains the chromosomes which controls the cell.

Given

Distance between two helium nuclei is 6nm

Find

The electrostatic repulsion between two helium nuclei.

Explanation:

The formula of electrostatic force is equal to F=k*Q1*Q2/d2

The electrostatic repulsion between two helium nuclei is

F=$9*10^{9} *(2*1.6*10^{-19} )^{2} / (6*10^{-9})^{2}$

=25.6 pN

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