Electric potential — definition V = −∫E·dl: Exercises

The geometry for the two-charge problems looks like this — keep it in mind:

Figure — Electric potential — definition V = −∫E·dl

Level 1 — Recognition

Recall Solution L1·Q1

WHAT: We ask how potentials combine. WHY it matters: $V$ is a scalar, built from a dot-product that was already collapsed into a single number. Answer: You add the plain numbers (with their $+ / -$ signs), never arrows. Superposition of potential is $V_{total} = \sum_{i} \frac{k q _{i}}{r _{i}}$ — ordinary algebraic addition. Direction lives only in $E = - \nabla V$ , not in $V$ .

Recall Solution L1·Q2

WHY V/m is legal: from $E = Δ V / d$ (uniform field), field = (volts) ÷ (metres) = V/m. And $1 V = 1 J/C = 1 N\cdotm/C$ , so $1 \frac{V}{m} = 1 \frac{N\cdotm/C}{m} = 1 \frac{N}{C} .$ V/m and N/C are the same unit. Both are correct.

Level 2 — Application

Recall Solution L2·Q1

Use $V = \frac{k q}{r}$ (derived by walking in radially from $\infty$ so $E ∥ d l$ ): $V = \frac{( 8.99 \times 1 0 ^{9} ) ( 5 \times 1 0 ^{- 9} )}{0.20} = 224.75 V .$ Sign check: positive charge ⇒ $V > 0$ , and it grows as $r$ shrinks. Correct.

Recall Solution L2·Q2

Uniform field ⇒ $E$ leaves the integral, giving $E = \frac{Δ V}{d}$ (see Capacitance and parallel-plate fields (E = ΔV/d)): $E = \frac{12}{4.0 \times 1 0 ^{- 3}} = 3000 V/m .$ Direction: $E$ points from the high-potential plate to the low-potential plate (field runs downhill in $V$ ).

Recall Solution L2·Q3

$E = - \nabla V$ — differentiate the scalar in each direction and flip sign (see Gradient operator ∇ and directional derivatives): $E_{x} = - \partial_{x} V = - 8 x, E_{y} = - \partial_{y} V = 3 z, E_{z} = - \partial_{z} V = 3 y .$ At $(1, 2, 1)$ : $E = (- 8, 3, 6)$ V/m. The additive constant $+ 7$ vanishes — only changes in $V$ produce field.

Level 3 — Analysis

Recall Solution L3·Q1

Work by field $= q (V_{A} - V_{B})$ (from Work–energy theorem in electrostatics): $W_{field} = (- 3 \times 1 0^{- 6}) (40 - 90) = (- 3 \times 1 0^{- 6}) (- 50) = + 1.5 \times 1 0^{- 4} J .$ Analysis: the number came out positive — the field helped the motion. Why? A negative charge is pushed toward higher potential (opposite to a positive charge), so moving to larger $V$ is "downhill" for it. The sign of $q$ decided everything.

Recall Solution L3·Q2

Add scalar contributions (superposition, no arrows). Both charges are $0.15$ m from $M$ :

= \frac{8.99\times10^{9}}{0.15}\left(4\times10^{-9} - 2\times10^{-9}\right). $$ $$ V_M = (5.993\times10^{10})(2\times10^{-9}) = 119.87\ \text{V}. $$ **Analysis:** the negative charge *subtracts*, so $V_M$ is smaller than the $+4$ nC alone would give. This is why $V$ can be zero or negative even where $\mathbf{E}\neq 0$.

Recall Solution L3·Q3

Let the point be at distance $x$ from $q_{1}$ , so distance $0.30 - x$ from $q_{2}$ . Set $V = 0$ :

\;\Rightarrow\; \frac{4}{x} = \frac{2}{0.30-x}. $$ Cross-multiply: $4(0.30 - x) = 2x \Rightarrow 1.20 = 6x \Rightarrow x = 0.20$ m. **Analysis:** $V=0$ here, but $\mathbf{E}\neq 0$ — potential is a *sum* that can cancel while the vector field does not. Zero potential ≠ zero field.

Level 4 — Synthesis

Recall Solution L4·Q1

(a) Outside ( $r > R$ ): field is point-like, so integrating in from $\infty$ gives $V (r) = - \int_{\infty}^{r} \frac{k Q}{r ^{'2}} d r^{'} = \frac{k Q}{r} .$ At $r = R$ the surface value is $V (R) = \frac{k Q}{R}$ .

(b) Inside ( $r < R$ ): here $E = 0$ , so walking further inward adds nothing to the integral: $V (r) = V (R) - \int_{R}^{r} = 0 E d r^{'} = V (R) = \frac{k Q}{R} (constant) .$ WHY constant: zero field means zero slope, so $V$ is flat inside — the whole interior is an equipotential.

Centre value: $V (0) = \frac{k Q}{R} = \frac{( 8.99 \times 1 0 ^{9} ) ( 8 \times 1 0 ^{- 9} )}{0.10} = 719.2 V .$

Recall Solution L4·Q2

All the potential energy converts to kinetic energy (from Work–energy theorem in electrostatics): $Δ KE = q (V_{start} - V_{end}) = (1.6 \times 1 0^{- 19}) (719.2 - 0) .$ $Δ KE = 1.15072 \times 1 0^{- 16} J .$ Sign check: positive charge, positive $V$ , moves to lower $V$ — energy is released, KE positive. Good.

Level 5 — Mastery

Recall Solution L5·Q1

Walk radially so $E \cdot d l = E d r$ :

= -\frac{\lambda}{2\pi\varepsilon_0}\ln\!\frac{r_b}{r_a}. $$ **Why infinity fails:** if $r_a$ or $r_b \to \infty$, $\ln(\infty)$ **diverges** — the integral has no finite value. So you *cannot* put $V=0$ at $\infty$ here; you pick a finite reference radius. Only the **difference** (a logarithm of a ratio) is physical. This is exactly the parent's warning about infinite distributions.

Recall Solution L5·Q2

$E_{r} = - \frac{d}{d r} (\frac{k q}{r}) = - (- \frac{k q}{r ^{2}}) = \frac{k q}{r ^{2}} .$ This is exactly Coulomb's field. WHY this matters: the two master relations $V = - \int E \cdot d l$ and $E = - \nabla V$ are perfect inverses — integrate the field to get $V$ , differentiate $V$ to get the field back. The minus signs cancel to give the correct outward-pointing field for $q > 0$ .

Recall Solution L5·Q3

False. Force comes from $E = - \nabla V$ (the slope), not from $V$ itself. At $x = 0.20$ m: both charges' fields point in the same direction on that segment (away from $+ q_{1}$ and toward $- q_{2}$ both point in $+ x$ ), so they add, not cancel: $E = \frac{k ( 4 \times 1 0 ^{- 9} )}{0.2 0 ^{2}} + \frac{k ( 2 \times 1 0 ^{- 9} )}{0.1 0 ^{2}} \neq = 0.$ $V = 0$ is a value; the gradient of $V$ there is large. Zero potential, strong field — the mastery lesson of this whole page.

Connections

Parent: Electric potential definition
Electric field E — definition and Coulomb's law
Potential energy of a charge configuration
Equipotential surfaces and why E ⟂ them
Gradient operator ∇ and directional derivatives
Work–energy theorem in electrostatics
Capacitance and parallel-plate fields (E = ΔV/d)