Question

what is the electric potential a distance of 2.7×10^(-15) m away from a proton? express your answer using two significant figures.
part b
what is the electric potential energy of a system that consists of two protons 2.7×10^(-15) m apart - as might occur inside a typical nucleus? express your answer using two significant figures.

Explanation:

Step1: Recall electric - potential formula

The electric potential due to a point charge $q$ at a distance $r$ from it is given by $V = \frac{kq}{r}$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$ and the charge of a proton $q = 1.6\times 10^{-19}\ C$, and $r = 2.7\times10^{-15}\ m$.
$V=\frac{(9\times 10^{9}\ N\cdot m^{2}/C^{2})\times(1.6\times 10^{-19}\ C)}{2.7\times10^{-15}\ m}$

Step2: Calculate the electric potential

$V=\frac{9\times1.6\times 10^{9 - 19}}{2.7\times10^{-15}}\ V=\frac{14.4\times10^{- 10}}{2.7\times10^{-15}}\ V = 5.33\times10^{5}\ V$. Rounding to two significant figures, $V = 5.3\times10^{5}\ V$.

Step3: Recall electric - potential - energy formula for two - charge system

The electric potential energy of a two - charge system is given by $PE=\frac{kq_1q_2}{r}$. Here, $q_1 = q_2=1.6\times 10^{-19}\ C$, $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, and $r = 2.7\times10^{-15}\ m$.
$PE=\frac{(9\times 10^{9}\ N\cdot m^{2}/C^{2})\times(1.6\times 10^{-19}\ C)\times(1.6\times 10^{-19}\ C)}{2.7\times10^{-15}\ m}$

Step4: Calculate the electric potential energy

$PE=\frac{9\times1.6\times1.6\times10^{9-19 - 19}}{2.7\times10^{-15}}\ J=\frac{23.04\times10^{-29}}{2.7\times10^{-15}}\ J=8.53\times10^{-14}\ J$. Rounding to two significant figures, $PE = 8.5\times10^{-14}\ J$.

Answer:

Part A: $V = 5.3\times10^{5}\ V$
Part B: $PE = 8.5\times10^{-14}\ J$