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Question

Physics Question on Electric Field

The nucleus of helium atom contains two protons that are separated by distance 3.0×1015m3.0 \times 10^{-15}\, m. The magnitude of the electrostatic force that each proton exerts on the other is

A

20.6N20.6 \,N

B

25.6N25.6 \,N

C

15.6N15.6 \,N

D

12.6N12.6 \,N

Answer

25.6N25.6 \,N

Explanation

Solution

Charge of proton is qp=1.6×1019Cq_p = 1.6 \times 10^{-19}\, C Distance between the protons is, r=3×1015mr = 3 \times 10^{-15}\, m The magnitude of electrostatic force between protons is Fe=qpqp4πε0r2F_{e}=\frac{q_{p}\,q_{p}}{4\pi\varepsilon_{0}r^{2}} =9×109×1.6×1019×1.6×1019(3×1015)2=\frac{9\times10^{9}\times1.6\times10^{-19}\times1.6\times10^{-19}}{\left(3\times10^{-15}\right)^{2}} =25.6N=25.6\,N