Question

The electric potential at the surface of an atomic nucleus $\left( Z=50 \right)$ of radius of $9\times {{10}^{-15}}m$ is:
(A) 80V
(B) $8\times {{10}^{6}}V$
(C) 9V
(D) $9\times {{10}^{5}}V$

Answer 1

Hint: We know that the voltage difference between any two points in a circuit is known as Potential Difference and it is this potential difference which makes current flow. Unlike current which flows around a closed electrical circuit in the form of electrical charge, potential difference does not move or flow when it is applied. Electrical potential difference is the difference in the amount of potential energy a particle has due to its position between two locations in an electric field. This important concept provides the basis for understanding electric circuits. Based on this concept we have to solve this question.

Complete step-by-step answer:
According to the Faraday’s Law states that the instantaneous EMF (voltage) induced in a circuit equals the rate of change of magnetic flux through the circuit. The value of an alternating quantity at a particular instant is called instantaneous value. The graph of instantaneous values is plotted of an alternating quantity plotted against time is called waveform. Based on this concept we have to solve this question.
Total charge in the nucleus is $q=Ze$
The electric potential at the surface of the nucleus is $V=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{r}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Ze}{r}$
$Or\,\,V=\left( 9\times {{10}^{9}} \right)\times \dfrac{50\times 1.6\times {{10}^{-19}}}{9\times {{10}^{-15}}}=8\times {{10}^{6}}V$

Hence, the correct answer is Option B.

Note: We know that the electric potential, or voltage, is the difference in potential energy per unit charge between two locations in an electric field. When we talked about electric fields, we chose a location and then asked what the electric force would do to an imaginary positively charged particle if we put one there.
We know that the electrical potential energy is positive if the two charges are of the same type, either positive or negative, and negative if the two charges are of opposite types. This makes sense if we think of the change in the potential energy as we bring the two charges closer or move them farther apart.