Question

What is the electric potential energy of a system that consists of two protons 2.1×10−15 m apart-as might occur inside a typical nucleus? Express your answer using two significant figures.

Answer 1

Answer:

$1.1\times 10^{-12}\ J$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

q = Charge of electron = $1.6\times 10^{-19}\ C$

r = Distance of charges = $2.1\times 10^{-15}\ m$

Electric potential energy is given by

$V=\dfrac{kqq}{r}\\\Rightarrow V=\dfrac{8.99\times 10^9\times (1.6\times 10^{-19})^2}{2.1\times 10^{-15}}\\\Rightarrow V=1.0959238095\times 10^{-13}\ V$

The electric potential energy of the system is $1.1\times 10^{-13}\ J$