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Question: An electron moves from a point where the electric potential is \(20{\text{ }}V\) to a point where th...

An electron moves from a point where the electric potential is 20 V20{\text{ }}V to a point where the electric potential is 120 V120{\text{ }}V. If the electronic charge is 1.6 × 10−19 C1.6{\text{ }} \times {\text{ }}{10^{ - 19}}{\text{ }}C, the energy gained by the electron is
A. 1.92 × 10−21 J 1.92{\text{ }} \times {\text{ }}{10^{ - 21}}{\text{ }}J\:
B. 1.6 × 10−21 J 1.6{\text{ }} \times {\text{ }}{10^{ - 21}}{\text{ }}J\:
C. 1.92 × 10−17 J 1.92{\text{ }} \times {\text{ }}{10^{ - 17}}{\text{ }}J\:
D. 1.6 × 10−17 J 1.6{\text{ }} \times {\text{ }}{10^{ - 17}}{\text{ }}J\:

Explanation

Solution

We will use the formulae for potential and potential energy. Then substitute the values at their respective positions. Finally, we will calculate the value of the energy. Electric potential difference can be defined as the work done in moving the test charge from one point to the other within the area of concern.

Formulae used:
ΔV = kqr\Delta V{\text{ }} = {\text{ }}\dfrac{{kq}}{r}
⇒U = kq2r\Rightarrow U{\text{ }} = {\text{ }}\dfrac{{k{q^2}}}{r}

Complete step by step answer:
We know that the formula for electric potential is
ΔV = kqr\Delta V{\text{ }} = {\text{ }}\dfrac{{kq}}{r}
Also, we know that the formula for electric potential energy is
U = kq2rU{\text{ }} = {\text{ }}\dfrac{{k{q^2}}}{r}
Now, if we compare the equations of both the parameters, we get
U = ΔV × qU{\text{ }} = {\text{ }}\Delta V{\text{ }} \times {\text{ }}q
Now, the potential difference for the given situation
ΔV = (120 − 20) V = 100 V\Delta V{\text{ }} = {\text{ }}(120{\text{ }} - {\text{ }}20){\text{ }}V{\text{ }} = {\text{ }}100{\text{ }}V
And the given value of charge is q = 1.6 × 10−19 Cq{\text{ }} = {\text{ }}1.6{\text{ }} \times {\text{ }}{10^{ - 19}}{\text{ }}C
Putting in the values in the equation of potential energy, we get
U = 100 V × 1.6 × 10−19 CU{\text{ }} = {\text{ }}100{\text{ }}V{\text{ }} \times {\text{ }}1.6{\text{ }} \times {\text{ }}{10^{ - 19}}{\text{ }}C
Thus, the final answer turns out to be
∴U = 1.6 × 10−17 J\therefore U{\text{ }} = {\text{ }}1.6{\text{ }} \times {\text{ }}{10^{ - 17}}{\text{ }}J

Hence, the correct answer is D.

Additional Information: Electric potential can be defined as the work done in bringing a test charge from infinity to a point in the region of concern. While potential energy is the energy required to move a charge away from an electric field.

Note: Students should always try to manipulate the known formulae otherwise they might fall into a situation of extreme dilemma and might lead them to some clumsy calculation which also might not serve the purpose of solving the given problem. They should always try to manipulate the formulae such that they are only required to use the given parameters in the question. Students should always keep in mind that the answers should generally be calculated in scientific notations.