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3 years ago

What is the electric potential v due to the nucleus of hydrogen at a distance of 5.292×10−11 m ?

Physics

1 answer:

viktelen [127]3 years ago

5 0

Answer: 27.21 V

Explanation:

The electric potential $V_{E}$ due to a point charge is expressed as:

$V_{E}=k\frac{q}{r}$

Where:

$k=9(10)^{9}\frac{Nm^{2}}{C^{2}}$ is the electric constant

$q=1.6(10)^{-19}C$ is the electric charge of the hydrogen nucleus, which is positive

$r=5.292(10)^{-11}m$ is the distance

Rewritting the equation with the known values:

$V_{E}=9(10)^{9}\frac{Nm^{2}}{C^{2}}\frac{1.6(10)^{-19}C}{5.292(10)^{-11}m}$

Finally:

$V_{E}=27.21 V$

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Replacing:

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