Electric Field & Potential Relationship - Physics 2

1. Which of the following statements about the electric field in that region is necessarily true?
   Answer: The electric field is zero throughout the region, because a constant potential means zero potential gradient, and $\vec{E} = -\nabla V$ requires a nonzero gradient to produce a nonzero field. Whenever you see a question about electric fields and potential, your first instinct should be to recall the fundamental relationship $\vec{E} = -\nabla V$. This single equation tells you everything: the electric field is the negative gradient of the electric potential. If the potential is constant throughout an entire volume — meaning it has the same value $V_0$ at every point — then every partial derivative of $V$ is zero: $\frac{\partial V}{\partial x} = \frac{\partial V}{\partial y} = \frac{\partial V}{\partial z} = 0$. Therefore, $\vec{E} = -\nabla V = \vec{0}$ throughout the entire region. That's exactly what A states, making it the correct answer. B is wrong because it invents a special exception at the boundary while claiming the interior field could be nonzero — this contradicts $\vec{E} = -\nabla V$ directly. There is no mechanism by which a field can exist "along" an equipotential surface; field lines are always perpendicular to equipotentials, not parallel to them. C is wrong on two counts: if the potential is constant, the field cannot be nonzero, and a constant potential in a volume doesn't produce "parallel plane" equipotentials — the entire volume is one equipotential, so the gradient is zero everywhere. D is wrong because the gradient of a constant function is unambiguously zero — not undefined. No additional boundary conditions are needed; the math resolves completely. Study tip: Treat $\vec{E} = -\nabla V$ as a checklist: zero gradient → zero field, always. Questions that suggest a nonzero field can somehow hide inside a region of constant potential are always traps.
2. What is the radial component of the electric field, and how does it depend on $r$?
   Answer: $E_r = +\frac{2A}{r^3}$, directed radially outward, because $E_r = -dV/dr = -(-2A/r^3) = +2A/r^3$, and since $A > 0$ the result is positive, indicating the field points away from the origin. Whenever you see a question linking electric potential to electric field, your anchor formula is $E_r = -\frac{dV}{dr}$. The negative sign is not optional — it encodes the physical fact that the field points in the direction of decreasing potential. Here, $V(r) = Ar^{-2}$, so differentiate with respect to $r$: $\frac{dV}{dr} = -2Ar^{-3} = -\frac{2A}{r^3}$ Now apply the definition: $E_r = -\frac{dV}{dr} = -\left(-\frac{2A}{r^3}\right) = +\frac{2A}{r^3}$ Because $A > 0$, the result is positive, meaning the field points radially outward from the origin. That confirms choice B. Choice A commits two errors: it invents the shortcut $E_r = V/r$ (which has no physical basis) and then incorrectly labels the direction as inward despite getting a negative sign through faulty reasoning. Choice C reaches the right magnitude but flips the sign. The reasoning — "potential decreases, so the derivative must be negative" — confuses $dV/dr$ with $E_r$. Yes, $dV/dr$ is negative here, but multiplying by the negative sign in the formula gives a positive $E_r$, so the field actually points outward. Choice D gets the direction right but drops the factor of 2 that comes from the power-rule derivative of $r^{-2}$. Study tip: Always apply the power rule carefully when differentiating potentials of the form $Ar^n$ — the exponent multiplies down and decreases by one. And never skip the leading negative sign in $E_r = -dV/dr$; it's the most common single-step mistake on field-from-potential problems.
3. In a source-free region of space (no charges present), the electric potential satisfies Laplace's equation $\nabla^2 V = 0$. A student asserts: 'Because Laplace's equation forbids local maxima and minima of $V$ in the interior, the gradient of $V$ — and therefore $\vec{E}$ — can never be zero inside the region.' Which of the following best evaluates this reasoning?
   Answer: The reasoning is flawed in its conclusion: although Laplace's equation does forbid local maxima and minima, $\vec{E}$ can still be zero at isolated saddle points of $V$, which are critical points where $\nabla V = 0$ but which are not extrema, and are fully consistent with $\nabla^2 V = 0$. Whenever you see a question mixing Laplace's equation with electric field behavior, separate two distinct concepts: what Laplace's equation says about extrema of $V$, and what it says about zeros of $\nabla V$. The maximum principle for harmonic functions (solutions to $\nabla^2 V = 0$) is real and important: $V$ cannot have a strict local maximum or minimum in the interior of a source-free region. The student correctly identifies this principle. However, the student then makes a critical logical error — assuming that "no extrema" implies "no zeros of $\nabla V$." This leap is wrong. A zero of $\nabla V$ (i.e., $\vec{E} = 0$) does not require an extremum. It only requires a critical point, which includes saddle points. At a saddle point, $\nabla V = 0$, yet $V$ curves upward in some directions and downward in others — no extremum exists, and $\nabla^2 V = 0$ is perfectly satisfiable. A classic example is the potential near two equal positive charges: the midpoint is a saddle point where $\vec{E} = 0$, entirely consistent with Laplace's equation. This makes D correct. A is wrong because it accepts the student's flawed conclusion. Even the caveat about equal-potential boundaries doesn't rescue the core error. B is wrong in the opposite direction — it denies the maximum principle entirely, which is a well-established theorem for harmonic functions. C is wrong because it invents an asymmetry that doesn't exist: Laplace's equation forbids both maxima and minima equally. Remember: "no extrema" $\neq$ "no critical points." Saddle points are the loophole that appears repeatedly in electrostatics problems.
4. A conducting shell of radius $R$ carries a net charge $+Q$. Inside the shell ($r < R$), the electric field is zero. A student argues: 'Since $\vec{E} = -\nabla V = 0$ inside, the potential inside must be zero.' Which response best addresses this argument?
   Answer: The argument is incorrect: $\vec{E} = -\nabla V = 0$ implies that $V$ is constant inside the shell, not necessarily zero. The constant value equals the surface potential $V = kQ/R$, set by continuity with the exterior solution at the boundary. When you see a question linking electric field to potential inside a symmetric charge distribution, the key concept to test is whether "zero gradient" means "zero value." These are fundamentally different statements. The relationship $\vec{E} = -\nabla V = 0$ tells you that the rate of change of potential is zero — not that the potential itself is zero. Mathematically, if $\nabla V = 0$ everywhere in a region, then $V$ must be constant throughout that region. The value of that constant is determined by boundary conditions, not by the condition $\vec{E} = 0$ alone. Since the potential must be continuous at $r = R$, the interior constant must match the surface value: $V = kQ/R$. This makes B the correct answer. Choice A correctly identifies that the potential difference between any two interior points is zero (since $\int \vec{E} \cdot d\vec{\ell} = 0$), but then makes the critical error of concluding the potential value is zero. Zero potential difference means the potential is uniform — it says nothing about the absolute value. Choice C is physically and mathematically wrong: $\nabla V = 0$ implies constant potential, never a linear variation. A linear potential would produce a nonzero uniform field, contradicting $\vec{E} = 0$. Choice D confuses two separate ideas — while you can always choose a reference point, the "conventional choice" for bounded distributions sets $V = 0$ at infinity, which gives $V = kQ/R$ on the surface, not zero. A reliable study tip: always distinguish between zero derivative and zero value. On electrostatics problems, $\vec{E} = 0$ constrains how $V$ changes, but boundary conditions determine what $V$ actually equals.
5. Which of the following best evaluates the student's claim?
   Answer: The claim is incorrect because the electric field depends on the rate of change of potential with position, not on the value of the potential itself. The potential changes at a constant rate $2V_0/d$ throughout the gap, so the field is uniform and nonzero everywhere between the plates. Whenever you see a question linking electric potential and electric field, the critical concept to recall is that the electric field is not the potential itself — it is the spatial rate of change of potential. Mathematically, $E = -\frac{dV}{dx}$. This means the field depends on how steeply the potential varies from point to point, not on whatever numerical value the potential happens to have at a given location. In the parallel-plate setup described, the potential drops linearly from $+V_0$ to $-V_0$ across the gap of width $d$. That linear slope is constant: $\frac{\Delta V}{\Delta x} = \frac{2V_0}{d}$ everywhere between the plates, including at the midpoint. Therefore the electric field is uniform and equal to $\frac{2V_0}{d}$ throughout the entire gap — it never vanishes. Answer B correctly identifies this reasoning and is the right choice. Answer A is built on a fundamental misconception: that $E = V/d$ means zero potential implies zero field. This formula (more properly written $E = \Delta V / d$) involves the difference in potential across the gap, not the value at a single point. Zero potential at the midpoint says nothing about the field there. Answer C invents a false rule — that field strength is proportional to the absolute value of the local potential. This has no basis in electrostatics; the field depends on the gradient, not the magnitude, of $V$. Answer D is also wrong. Equipotential surfaces are always perpendicular to field lines — they never coincide with them. A zero-potential plane does not cause the field to vanish. Study tip: Any time a question mentions potential at a point, ask yourself: is this asking about the value of V, or its rate of change? The electric field only cares about the slope.
6. Using a finite-difference approximation, what is the best estimate of the x-component of the electric field at point Q?
   Answer: $E_x(Q) \approx +3$ V/m, in the $+x$ direction, using the central difference: $E_x \approx -(V_R - V_P)/(x_R - x_P) = -(4-10)/2 = +3$ V/m, since the potential decreases from P to R and the field points in the direction of decreasing potential. When a question asks for the electric field at an interior point given potentials at surrounding points, your instinct should be to use the central difference approximation — it's more accurate than a one-sided difference because it uses information from both sides of the point. The core relationship is $E_x = -\frac{dV}{dx}$, which in finite-difference form becomes $E_x(Q) \approx -\frac{V_R - V_P}{x_R - x_P}$. Plugging in: $E_x(Q) \approx -\frac{4 - 10}{2 - 0} = -\frac{-6}{2} = +3 \text{ V/m}$. This is positive, meaning the field points in the $+x$ direction — which makes physical sense, since the potential is dropping as you move toward R, and electric fields always point from high to low potential. Choice C captures this correctly. Choice A uses only the forward difference (Q to R), which is a less accurate one-sided estimate. It gives +6 V/m, not the best approximation available when data from both sides exists. Choice B performs the central difference calculation correctly but then misreads the sign — the result $-(4-10)/2 = +3$ V/m is positive, not negative. Labeling it $-3$ V/m in the $-x$ direction is a sign error. Choice D confuses "no potential difference between P and Q" with "no field at Q." The field at Q depends on the potential gradient across Q, not just one neighboring segment. Study tip: On finite-difference problems, always prefer the central difference when points exist on both sides, and carefully track the negative sign in $E_x = -dV/dx$ — it's the most common place students lose points.
7. Which of the following correctly describes both the direction and the $x$-dependence of the electric field component $E_x$ along this line?
   Answer: $E_x = +\alpha V_0 e^{-\alpha x}$, pointing in the $+x$ direction for all $x \geq 0$, because $E_x = -dV/dx = \alpha V_0 e^{-\alpha x} > 0$. The field is strongest at $x = 0$ and decays exponentially, reflecting the steeper potential gradient near the origin. Whenever you see a question connecting electric potential to electric field, your anchor relationship is $E_x = -\dfrac{dV}{dx}$. The negative sign is critical — it tells you the field points in the direction of decreasing potential, not increasing. Here, $V(x) = V_0 e^{-\alpha x}$, so taking the derivative: $\dfrac{dV}{dx} = -\alpha V_0 e^{-\alpha x}$. Applying the formula: $E_x = -\left(-\alpha V_0 e^{-\alpha x}\right) = +\alpha V_0 e^{-\alpha x}$. This is positive for all $x \geq 0$, confirming the field points in the $+x$ direction. The magnitude is largest at $x = 0$ (where the potential gradient is steepest) and decays exponentially — exactly what C describes. A gets the math backwards by claiming the field points toward higher potential. Fields always point toward lower potential (away from positive charges, toward negative). The two negatives — one from the formula, one from the derivative — combine to give a positive field, not negative. B correctly computes the sign but then claims the magnitude is constant at $\alpha V_0$. This ignores the $e^{-\alpha x}$ factor entirely. The exponential decays with position, so the field strength absolutely changes with $x$. D makes a calculus error, claiming $\dfrac{d}{dx}(e^{-\alpha x}) = e^{-\alpha x}$, which would only be true if $\alpha = -1$. The chain rule requires bringing down $-\alpha$, which cannot simply be "absorbed." Study tip: Always apply the chain rule carefully to exponentials, and remember the field points downhill in potential — never toward higher $V$.

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