Question:

The nucleus of helium atom contains two protons that are separated by distance $3.0 \times 10^{-15}\, m$. The magnitude of the electrostatic force that each proton exerts on the other is

Updated On: Jul 7, 2022
  • $20.6 \,N$
  • $25.6 \,N$
  • $15.6 \,N$
  • $12.6 \,N$
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The Correct Option is B

Solution and Explanation

Charge of proton is $q_p = 1.6 \times 10^{-19}\, C$ Distance between the protons is, $r = 3 \times 10^{-15}\, m$ The magnitude of electrostatic force between protons is $F_{e}=\frac{q_{p}\,q_{p}}{4\pi\varepsilon_{0}r^{2}}$ $=\frac{9\times10^{9}\times1.6\times10^{-19}\times1.6\times10^{-19}}{\left(3\times10^{-15}\right)^{2}}$ $=25.6\,N$
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Notes on Electric Field

Concepts Used:

Electric Field

Electric Field is the electric force experienced by a unit charge. 

The electric force is calculated using the coulomb's law, whose formula is:

\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)

While substituting q2 as 1, electric field becomes:

 \(E=k\dfrac{|q_{1}|}{r^{2}}\)

SI unit of Electric Field is V/m (Volt per meter).