Physics 2 Quiz: Electric Field And Potential Relationship

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Electric Field And Potential RelationshipQuestion 1 of 7

A physicist maps the electric potential along a straight line and finds that the potential varies as $V(x) = V_0 e^{-\alpha x}$ for $x \geq 0$ , where $V_0 > 0$ and $\alpha > 0$ are constants.

Which of the following correctly describes both the direction and the $x$ -dependence of the electric field component $E_x$ along this line?

E_x = -\alpha V_0 e^{-\alpha x}

, pointing in the

-x

direction for all

x \geq 0

, because the potential decreases with increasing

x

and the field must point toward higher potential, i.e., in the

-x

direction.

E_x = +\alpha V_0 e^{-\alpha x}

, pointing in the

+x

direction, with constant magnitude

\alpha V_0

for all

x \geq 0

, because the exponential function maintains a fixed amplitude that does not change with position.

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.

E_x = +V_0 e^{-\alpha x}

, pointing in the

+x

direction, because for an exponential function

dV/dx = V

, so

E_x = -dV/dx = -V = -V_0 e^{-\alpha x}

; taking the magnitude gives

V_0 e^{-\alpha x}

, with the factor of

\alpha

absorbed into the definition of the field.

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Physics 2 Quiz

Physics 2 Quiz: Electric Field And Potential Relationship

Practice Electric Field And Potential Relationship in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Electric Field And Potential Relationship, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A physicist maps the electric potential along a straight line and finds that the potential varies as $V(x) = V_0 e^{-\alpha x}$ for $x \geq 0$ , where $V_0 > 0$ and $\alpha > 0$ are constants.

Which of the following correctly describes both the direction and the $x$ -dependence of the electric field component $E_x$ along this line?

$E_x = -\alpha V_0 e^{-\alpha x}$ , pointing in the $-x$ direction for all $x \geq 0$ , because the potential decreases with increasing $x$ and the field must point toward higher potential, i.e., in the $-x$ direction.
$E_x = +\alpha V_0 e^{-\alpha x}$ , pointing in the $+x$ direction, with constant magnitude $\alpha V_0$ for all $x \geq 0$ , because the exponential function maintains a fixed amplitude that does not change with position.
$E_x = +\alpha V_0 e^{-\alpha x}$ , 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. (correct answer)
$E_x = +V_0 e^{-\alpha x}$ , pointing in the $+x$ direction, because for an exponential function $dV/dx = V$ , so $E_x = -dV/dx = -V = -V_0 e^{-\alpha x}$ ; taking the magnitude gives $V_0 e^{-\alpha x}$ , with the factor of $\alpha$ absorbed into the definition of the field.

Explanation: 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

Question 2

Three points — P, Q, and R — lie along the x-axis at $x = 0, 1,$ and $2$ m respectively. The electric potential at these points is measured to be $V_P = 10$ V, $V_Q = 10$ V, and $V_R = 4$ V.

Using a finite-difference approximation, what is the best estimate of the x-component of the electric field at point Q?

$E_x(Q) \approx +6$ V/m, in the $+x$ direction, using the forward difference from Q to R: $E_x \approx -(V_R - V_Q)/\Delta x = -(4-10)/1 = +6$ V/m.
$E_x(Q) \approx -3$ 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, where the negative sign indicates the field opposes the direction of increasing $x$ .
$E_x(Q) \approx +3$ 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. (correct answer)
$E_x(Q) \approx 0$ V/m, because $V_P = V_Q$ , indicating no potential gradient between P and Q, so the field at Q is zero.

Explanation: 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.

Question 3

A spherically symmetric charge distribution produces an electric potential $V(r) = \frac{A}{r^2}$ for $r > 0$ , where $A$ is a positive constant and $r$ is the radial distance from the center.

What is the radial component of the electric field, and how does it depend on $r$ ?

$E_r = -\frac{A}{r^3}$ , directed radially inward, because the field equals the potential divided by the radial distance, giving $E_r = V/r = A/r^3$ , and the negative sign reflects the inward direction.
$E_r = +\frac{2A}{r^3}$ , 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. (correct answer)
$E_r = -\frac{2A}{r^3}$ , directed radially inward, because the potential decreases as $r$ increases, so $-dV/dr$ must be negative, and the field therefore points toward the origin.
$E_r = +\frac{A}{r^3}$ , directed radially outward, because taking the negative derivative of $V = Ar^{-2}$ with respect to $r$ yields a positive quantity pointing away from the origin.

Explanation: 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.

Question 4

Two large parallel conducting plates are separated by a distance $d$ . The left plate is held at potential $V = +V_0$ and the right plate at $V = -V_0$ . A student claims: 'Because the potential changes uniformly from $+V_0$ to $-V_0$ , the electric field between the plates is zero at the midpoint where $V = 0$ .'

Which of the following best evaluates the student's claim?

The claim is correct because at the midpoint the potential is zero, and a region of zero potential must have zero electric field, since field and potential are proportional to each other by the relation $E = V/d$ .
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. (correct answer)
The claim is incorrect, but only because the field is not zero at the midpoint—it is actually largest there since the potential crosses zero, and the field is proportional to the absolute value of the potential at each location.
The claim is correct at the midpoint specifically, but the field is nonzero elsewhere. This is because equipotential surfaces and field lines coincide at $V = 0$ , making the field vanish only along that plane.

Explanation: 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

-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.

Question 5

In a certain region, the electric potential is observed to be constant throughout an extended volume — not just on a surface — with value $V = V_0$ .

Which of the following statements about the electric field in that region is necessarily true?

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. (correct answer)
The electric field is zero only on the boundary of the region; inside, the field could be nonzero and directed along equipotential surfaces, since field lines may curve within a uniform-potential volume.
The electric field is uniform and nonzero throughout the region, directed perpendicular to the equipotential surfaces, which are all parallel planes when the potential is constant in a volume.
The electric field magnitude is constant but its direction is undefined, because a constant potential provides no directional information about the gradient; additional boundary conditions are needed to determine $\vec{E}$ .

Explanation: 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.

Question 6

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?

The argument is correct: if the electric field is zero everywhere inside the shell, then by integrating $\vec{E} \cdot d\vec{\ell} = 0$ along any path inside, the potential difference between any two interior points is zero, which means the interior potential is zero.
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. (correct answer)
The argument is incorrect: $\vec{E} = 0$ inside means $\nabla V = 0$ , so $V$ varies linearly inside the shell. The linear variation approaches zero at the center, which is why the field vanishes there.
The argument is correct in principle but requires a reference point: the potential inside is zero only if we choose the zero of potential to be at the surface of the shell, which is the conventional choice for bounded charge distributions.

Explanation: 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.

Question 7

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?

The reasoning is correct: the maximum principle forbids all extrema of $V$ in the interior, so $\nabla V \neq 0$ everywhere inside, meaning $\vec{E} \neq 0$ at every interior point unless the entire region is bounded by equal-potential surfaces.
The reasoning is flawed from the start: Laplace's equation does not forbid local extrema of $V$ ; both maxima and minima are permitted in source-free regions, so the student's premise is incorrect.
The reasoning is partially correct: Laplace's equation forbids local maxima but permits local minima, because a minimum requires inward flux of $\vec{E}$ , which is consistent with $\nabla \cdot \vec{E} = 0$ in a charge-free region.
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 $\nabla^2 V = 0$ . (correct answer)

Explanation: 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.