What is the potential at a distance of 0.0529 nm from a proton? a.13.6 no b.-13.6...

Question

Question

What is the potential at a distance of 0.0529 nm from a proton?

a.13.6 no

b.-13.6 no

c.27.2 V

d. -27.2 no

e.9.11 pC

Answer 1

Answer #1

The electric potential from a distance r for a charge q is given by the equation is

$v = q/(4*pi*\epsilon *r)$

Here $q = 1.6*10^{-19}$ and $\epsilon = 8.85*10^{-12}$ and r = 0.0529 nm

By substituting the values in the equation we will get

v = 27.2 V

Answer 2

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