Problem Statement
Two point charges $q_1$ and $q_2$ are placed at known positions. Find the electric potential at a specified field point P.
Given Information
- Charge $q_1$ at position $r_1$ from P
- Charge $q_2$ at position $r_2$ from P
- Coulomb constant $k = 9\times10^9\,\text{N m}^2\text{C}^{-2}$
Physical Concepts & Formulas
Electric potential obeys the principle of superposition and, crucially, it is a scalar — so potentials from multiple sources add algebraically, not vectorially. This is the primary advantage of working with potential rather than field. The potential at any point due to a system of charges is simply the algebraic sum of the individual potentials, with careful attention to signs: positive charges contribute positive potential, negative charges contribute negative potential.
- $V = \dfrac{kq}{r}$ — potential due to single point charge
- $V_{\text{total}} = V_1 + V_2 = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}$ — superposition
- Work to bring test charge $q_0$: $W = q_0(V_f – V_i)$
Step-by-Step Solution
Step 1 — Identify distances: Compute $r_1$ and $r_2$ (distances from each charge to P) using geometry or the given values.
Step 2 — Apply superposition:
$$V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$$
Step 3 — Substitute values: Insert numerical values of $q_1$, $q_2$, $r_1$, $r_2$ with correct signs.
Step 4 — Combine: Add the two scalar contributions (watch signs for negative charges).
Worked Calculation
Substituting all values with units:
$$V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2} = k\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right)$$
Answer
$$\boxed{V = k\!\left(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2}\right)}$$
Physical Interpretation
If the result is positive, the field point is dominated by the positive charge(s) and a positive test charge would need to do work against the field to move away. If negative, the reverse is true. The sign and magnitude both carry physical meaning: $V$ is the work per unit charge needed to bring a test charge from infinity.
Leave a Reply