An electric field does 1.50 × 103 eV of work on a carbon nucleus of charge9.61 × 10−19 C. Find the change in the nucleus’i. electric potential.ii. electric potential energy in joules.
Question
An electric field does 1.50 × 103 eV of work on a carbon nucleus of charge9.61 × 10−19 C. Find the change in the nucleus’i. electric potential.ii. electric potential energy in joules.
Solution
i. Electric potential (V) is defined as the work done (W) per unit charge (Q). It can be calculated using the formula:
V = W/Q
Given: W = 1.50 × 10^3 eV = 1.50 × 10^3 * 1.6 × 10^-19 J (since 1 eV = 1.6 × 10^-19 J) Q = 9.61 × 10^-19 C
Substituting these values into the formula, we get:
V = (1.50 × 10^3 * 1.6 × 10^-19 J) / (9.61 × 10^-19 C) = 0.25 V
ii. Electric potential energy (U) is the work done by an external agent in moving a charge from a reference point to a specific point inside the field without producing an acceleration. It can be calculated using the formula:
U = QV
Substituting the given values into the formula, we get:
U = (9.61 × 10^-19 C) * (0.25 V) = 2.40 × 10^-19 J
So, the change in the nucleus' electric potential is 0.25 V and the change in the nucleus' electric potential energy is 2.40 × 10^-19 J.
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