Physics 2 Quiz: Electric Potential From Point Charges

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Electric Potential From Point ChargesQuestion 1 of 9

Two point charges, $+Q$ and $-Q$ , are separated by a distance $2a$ . A student claims that there exists a spherical surface centered on the midpoint between the charges on which the electric potential is everywhere zero. A second student claims that there exists a plane surface — perpendicular to the line joining the charges and passing through the midpoint — on which the potential is everywhere zero. Which student is correct, and why?

Only the first student is correct; the locus of zero potential for an electric dipole forms a sphere centered at the midpoint because the

1/r

dependence of each charge's potential guarantees equal magnitudes on that surface.

Only the second student is correct; every point on the perpendicular bisector plane is equidistant from

+Q

and

-Q

, so the potentials

+kQ/r

and

-kQ/r

cancel exactly, giving

V = 0

everywhere on that plane.

Both students are correct; the zero-potential surface is simultaneously a sphere and a plane because for equal and opposite charges these two geometric descriptions are equivalent by symmetry.

Neither student is correct; for a true electric dipole the zero-potential surface is an oblate ellipsoid, not a sphere or a plane, because the

1/r

potential falls off differently along the axis versus perpendicular to it.

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Physics 2 Quiz

Physics 2 Quiz: Electric Potential From Point Charges

Practice Electric Potential From Point Charges 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 Potential From Point Charges, 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

Only the first student is correct; the locus of zero potential for an electric dipole forms a sphere centered at the midpoint because the $1/r$ dependence of each charge's potential guarantees equal magnitudes on that surface.
Only the second student is correct; every point on the perpendicular bisector plane is equidistant from $+Q$ and $-Q$ , so the potentials $+kQ/r$ and $-kQ/r$ cancel exactly, giving $V = 0$ everywhere on that plane. (correct answer)
Both students are correct; the zero-potential surface is simultaneously a sphere and a plane because for equal and opposite charges these two geometric descriptions are equivalent by symmetry.
Neither student is correct; for a true electric dipole the zero-potential surface is an oblate ellipsoid, not a sphere or a plane, because the $1/r$ potential falls off differently along the axis versus perpendicular to it.

Explanation: When a question asks about zero-potential surfaces, your first instinct should be to write out the superposition principle: the total potential at any point is the scalar sum of contributions from each charge. For a point at distance

r_1

from

+Q

and

r_2

from

-Q

, the total potential is:

V = \frac{kQ}{r_1} + \frac{k(-Q)}{r_2} = kQ\left(\frac{1}{r_1} - \frac{1}{r_2}\right)

This equals zero only when

r_1 = r_2

— that is, when the point is equidistant from both charges. The geometric locus of all points equidistant from two fixed points in three dimensions is exactly the perpendicular bisecting plane between them. Every point on that plane satisfies

r_1 = r_2

, so

+kQ/r_1

and

-kQ/r_2

cancel perfectly. This confirms that B is correct. A is wrong because a sphere centered at the midpoint does not guarantee equal distances to both charges — a point on that sphere closer to

+Q

along the axis has

r_1 \neq r_2

, so the potentials don't cancel. The

1/r

dependence argument is misapplied here. C is wrong because a plane and a sphere are geometrically distinct surfaces. Only the plane satisfies

r_1 = r_2

everywhere; claiming they're equivalent "by symmetry" is simply false. D is wrong because the zero-potential surface for two equal and opposite point charges is exactly a plane — not an ellipsoid. That distractor confuses the potential geometry with field-line geometry or dipole approximation concepts. Study tip: For any two-charge system, zero potential means equal distances, and equal distances from two points always defines a plane — not a curved surface.

Question 2

Two point charges, $+3q$ and $-q$ , are separated by a distance $d$ .

At how many distinct points on the line passing through both charges (including points outside the segment between them) does the electric potential equal zero?

One point only, located between the two charges, where the contributions from $+3q$ and $-q$ cancel despite the charge ratio being 3:1.
Two points: one between the charges where the $+3q$ and $-q$ potentials cancel, and one outside the segment on the side of the $-q$ charge where the greater distance from $+3q$ reduces its contribution sufficiently. (correct answer)
Two points: one between the charges and one outside the segment on the side of the $+3q$ charge, because the larger positive charge dominates only at close range and its potential drops below the $-q$ contribution at large distances on that side.
No finite points, because the $+3q$ charge always produces a larger positive potential than the $-q$ charge produces a negative potential at any finite location on the line, so the net potential is always positive.

Explanation: When working with electric potential along the line connecting two point charges, remember that potential is a scalar quantity:

V = kq/r

. Unlike electric field, there's no direction to worry about — you simply add the signed contributions algebraically and set the sum to zero. For charges

+3q

and

-q

separated by distance

d

, you need to find where

\frac{k(+3q)}{r_1} + \frac{k(-q)}{r_2} = 0

, which simplifies to

\frac{3}{r_1} = \frac{1}{r_2}

, or

r_2 = \frac{r_1}{3}

. Between the charges, let

r_1 = x

and

r_2 = d - x

. Setting

3(d-x) = x

gives

x = \frac{3d}{4}

— a valid point between them. Outside the segment on the side of $-q$ , let the point be distance

r_1

from

+3q

and

r_2

from

-q

, with

r_1 = r_2 + d

. Substituting yields another valid solution. This confirms answer B is correct. Choice A is wrong because it claims only one zero exists, missing the exterior solution entirely. Choice C is wrong in its geometry — the exterior zero point lies on the

-q

side, not the

+3q

side. Intuitively, you need to be closer to the smaller-magnitude negative charge for it to "keep up" with the larger positive charge. Choice D is wrong because it ignores the

1/r

dependence; at the correct distances, magnitudes do balance. Study tip: Always check both regions — between the charges and outside on each side — when hunting for zero-potential points. The exterior solution is frequently overlooked and frequently tested.

Question 3

A student calculates the electric potential at a point P due to two point charges $+Q$ at position $A$ and $-2Q$ at position $B$ . Point P is such that $|PA| = r$ and $|PB| = 2r$ .

The student correctly finds the potential at P, then argues: 'Since the total potential at P is negative, the electric field at P must point toward a region of higher (less negative) potential, which is the direction away from charge $-2Q$ .' Is this reasoning valid, and what is the potential at P?

The potential calculation is correct at $V_P = -kQ/2r$ (net negative), but the field-direction reasoning is flawed because the electric field direction is determined by the gradient of the total potential, not by the sign of the potential at a single point.
The potential at P is $V_P = -kQ/(2r)$ (net negative), and the student's field-direction reasoning is valid because by convention the electric field always points from regions of higher to lower potential, which is away from the negative charge.
The potential at P is $V_P = kQ/r - kQ/r = 0$ after simplification, so the student's premise fails. Furthermore, a zero potential at a point does imply zero electric field there by Gauss's law applied locally.
The potential at P is $V_P = kQ/r - 2kQ/(2r) = 0$ , so the student's premise is incorrect; V = 0 at P, and the electric field direction cannot be inferred from a zero potential. (correct answer)

Explanation: When dealing with superposition of electric potentials, your first job is always to compute the scalar sum carefully before drawing any conclusions about field direction. The potential at P is the algebraic sum of contributions from both charges. The charge

+Q

sits at distance

r

from P, and

-2Q

sits at distance

2r

V_P = \frac{kQ}{r} + \frac{k(-2Q)}{2r} = \frac{kQ}{r} - \frac{kQ}{r} = 0

So the total potential at P is exactly zero — the student's entire argument collapses at the premise. This confirms D as correct: the net potential vanishes, making the claim that "the potential is negative" simply wrong. A is tempting because it correctly identifies the flaw in field-direction reasoning, but it gets the arithmetic wrong. It states

V_P = -kQ/2r

, which would require the two terms to not cancel — they do cancel exactly. B doubles down on both errors: it accepts the incorrect negative potential and endorses the flawed reasoning. The electric field points in the direction of steepest decrease of potential (i.e.,

\vec{E} = -\nabla V

). You cannot infer field direction from the sign of the potential at one isolated point — you need to know the potential in a neighborhood around that point. C correctly finds

V_P = 0

but then introduces a false claim: zero potential at a point does not imply zero electric field. A field can exist where potential happens to be zero (think of the midpoint between equal opposite charges). Study tip: Always separate two concepts: potential is a scalar value at a point; electric field requires the spatial gradient. Never infer field direction from the sign or magnitude of potential at a single point alone.

Question 4

Three point charges are arranged along the x-axis: a charge of $+2q$ at $x = -d$ , a charge of $-q$ at $x = 0$ , and a charge of $+2q$ at $x = +d$ .

At what location(s) on the x-axis, if any, is the electric potential equal to zero (other than at infinity)?

At $x = 0$ only, because the two $+2q$ charges contribute equal and opposite potentials that cancel, leaving only the $-q$ contribution, which is nonzero there.
At two symmetric points $x \approx \pm 0.24d$ , found by setting the scalar sum of all three potential contributions equal to zero and solving the resulting quadratic equation. (correct answer)
Nowhere on the finite x-axis; because all three charges are on the x-axis and the two positive charges dominate, the net potential is always positive between the charges and never reaches zero.
At $x = \pm d/2$ , obtained by balancing the potential from each $+2q$ charge against the potential from the $-q$ charge at the midpoint between each pair.

Explanation: When finding where electric potential equals zero, remember that potential is a scalar, not a vector. You simply add up the contributions

V = \frac{kq}{r}

from each charge algebraically — no direction components needed. Setting up the equation at position

x

(between the charges, for symmetry), the total potential from all three charges must sum to zero:

\frac{k(+2q)}{|x+d|} + \frac{k(-q)}{|x|} + \frac{k(+2q)}{|x-d|} = 0

By the symmetric arrangement, solutions must come in

\pm

pairs. Focusing on the region

0 < x < d

and simplifying, you get a quadratic equation whose solutions yield

x \approx \pm 0.24d

, confirming answer B is correct. Answer A contains a fundamental misconception — the two

+2q

charges don't produce "opposite" potentials anywhere on the axis. Both are positive charges, so both contribute positive potential values everywhere. They cannot cancel each other. Answer C is wrong because it confuses electric field reasoning (where vector cancellation matters) with electric potential reasoning. The negative charge's contribution is large enough near certain points to bring the total scalar sum to zero. Answer D applies faulty logic — it tries to "balance" potential from individual charge pairs independently rather than summing all three contributions simultaneously. There's no shortcut here; all charges contribute at every point. The key study tip: potential is a scalar sum. When a question asks where

V = 0

, write one equation with all charges included, then solve it — never treat the charges in isolated pairs.

Question 5

An isolated conducting sphere of radius $R$ carries a net charge $+Q$ . A point charge $+q$ is held fixed at a distance $r > R$ from the center of the sphere. Applying the superposition principle to compute the electric potential at a point P on the surface of the sphere, which statement is correct?

$V_{\text{surface}} = \frac{kQ}{R} + \frac{kq}{r}$ , because the sphere contributes $kQ/R$ at its surface and the external charge contributes $kq/r$ , evaluated at the sphere's center as a standard near-field approximation.
$V_{\text{surface}} = \frac{kQ}{R} + \frac{kq}{r - R}$ , because potential must be evaluated at the nearest point on the conductor's surface, which is at distance $r - R$ from the external charge.
$V_{\text{surface}} = \frac{k(Q+q)}{R}$ , because the conducting sphere and the external point charge form a single system whose total charge $Q + q$ determines the surface potential.
$V_{\text{surface}} = \frac{kQ}{R} + \frac{kq}{d}$ , where $d$ is the distance from the external charge to the specific surface point P being evaluated; this value varies across the sphere's surface since different surface points are at different distances from $+q$ . (correct answer)

Explanation: When dealing with electric potential and the superposition principle, remember that potential is a scalar quantity — you simply add contributions from each source. The key is evaluating each contribution at the correct distance to the specific point in question. For a conducting sphere carrying charge

+Q

, the potential it produces at any point on or outside its surface behaves as if all charge were concentrated at the center, giving

kQ/R

at the surface. The external point charge

+q

contributes

kq/d

, where

d

is the distance from

+q

to the specific point P being evaluated. Because the sphere's surface is curved and

+q

sits off to one side, different surface points sit at different distances from

+q

. This means the total surface potential varies across the surface — and choice D correctly captures this:

V_{\text{surface}} = \frac{kQ}{R} + \frac{kq}{d}

, with

d

depending on which point P you choose. Choice A incorrectly evaluates the external charge's contribution at

r

, the distance to the sphere's center — that would only be valid if P were located at the center, not on the surface. Choice B uses

r - R

as if P were always the nearest surface point to

+q

; this applies only to one specific point on the sphere, not a general surface point. Choice C incorrectly treats

+q

as part of the sphere itself — the external charge is not on or inside the conductor, so you cannot simply add

Q + q

and divide by

R

. Your takeaway: whenever superposition involves an off-center external source, always identify the actual distance from that source to the exact field point — never default to a convenient approximation like the center-to-center distance.

Question 6

A charge $+Q$ is at position $(0, a)$ and a charge $-Q$ is at position $(0, -a)$ , forming an electric dipole along the y-axis. A third charge $+Q$ is placed at position $(b, 0)$ . What is the total electric potential at the origin due to all three charges?

$V = \frac{kQ}{b}$ , because the origin is equidistant from the two dipole charges so their contributions cancel exactly, leaving only the potential from the charge at $(b, 0)$ . (correct answer)
$V = \frac{kQ}{a} + \frac{kQ}{b}$ , because the $+Q$ dipole charge at distance $a$ contributes $kQ/a$ and the $-Q$ charge contributes a negative term that is smaller in magnitude, with the net dipole contribution adding to $kQ/b$ .
$V = \frac{kQ}{b} + \frac{kQa^2}{b^3}$ , because the dipole produces a nonzero higher-order correction at the origin that combines with the $kQ/b$ term from the third charge.
$V = 0$ , because the origin lies on the perpendicular bisector of the dipole making its net potential zero there, and the third charge at distance $b$ is assumed to be so far away that its contribution is negligible.

Explanation: When you see a question involving electric potential from multiple charges, remember one crucial fact: electric potential is a scalar, not a vector. You simply add up the individual contributions

V = kq/r

from each charge algebraically, with no direction to worry about. For the dipole charges, the

+Q

sits at

(0, a)

and the

-Q

sits at

(0, -a)

. Both are exactly a distance

a

from the origin. Their contributions are

+kQ/a

and

-kQ/a

, which cancel perfectly to zero. The third charge

+Q

(b, 0)

is a distance

b

from the origin, contributing

+kQ/b

. The total is therefore

V = kQ/b

, making A correct. B is wrong because it incorrectly claims the dipole contributions don't fully cancel — they do, since both charges are equidistant from the origin. There's no "smaller magnitude" discrepancy; the distances are identical. C confuses electric potential with electric field. Higher-order multipole corrections appear in the field or in the potential evaluated far from the dipole, not at the origin itself. At the origin, the exact cancellation is complete with no correction term needed. D correctly identifies that the dipole contribution is zero, but then wrongly dismisses the third charge. No approximation is needed —

kQ/b

is a real, finite contribution regardless of how large

b

is. Study tip: Always distinguish between electric potential (scalar, uses

kq/r

) and electric field (vector, requires direction). Dipole potentials cancel on the perpendicular bisector every time — no exceptions, no corrections.

Question 7

Consider two concentric thin spherical shells. The inner shell has radius $R_1$ and carries charge $+Q$ . The outer shell has radius $R_2 > R_1$ and carries charge $-Q$ .

Using the superposition principle and the shell theorem, what is the electric potential at a point P located at radius $r$ with $R_1 < r < R_2$ (between the shells), and at a point S located at radius $r' > R_2$ (outside both shells)?

At P: $V_P = \dfrac{kQ}{r}$ ; at S: $V_S = 0$ , because outside both shells the charges $+Q$ and $-Q$ cancel, while between the shells only the inner shell's contribution is present as a point-charge potential.
At P: $V_P = \dfrac{kQ}{R_1}$ ; at S: $V_S = 0$ , because the inner shell's potential is constant throughout its interior and on its surface at $kQ/R_1$ , and the outer shell's contributions cancel outside.
At P: $V_P = \dfrac{kQ}{r} - \dfrac{kQ}{R_2}$ ; at S: $V_S = 0$ , because between the shells the inner shell contributes $kQ/r$ (P is outside it) and the outer shell contributes a constant $-kQ/R_2$ (P is inside it), while outside both shells the net charge is zero. (correct answer)
At P: $V_P = \dfrac{kQ}{R_1} - \dfrac{kQ}{R_2}$ ; at S: $V_S = 0$ , because both shells contribute constant interior potentials: $kQ/R_1$ from the inner shell and $-kQ/R_2$ from the outer shell, which apply uniformly throughout the interior of each shell.

Explanation: When dealing with concentric charged shells, your go-to tools are the shell theorem and superposition: treat each shell independently, then add their potentials. The shell theorem tells you that a uniformly charged shell looks like a point charge outside it, but contributes a constant potential equal to

kq/R

(its surface value) at every point inside it. For point P between the shells (

R_1 < r < R_2

): P is outside the inner shell, so it contributes

+kQ/r

. P is inside the outer shell, so the outer shell contributes a uniform

-kQ/R_2

throughout its interior. By superposition,

V_P = \dfrac{kQ}{r} - \dfrac{kQ}{R_2}

. For point S outside both shells (

r' > R_2

): both shells act like point charges at the origin, giving

+kQ/r' - kQ/r' = 0

. This confirms C is correct. Answer A correctly identifies the inner shell's contribution as

kQ/r

, but it forgets that the outer shell still contributes

-kQ/R_2

inside it — the outer shell's potential doesn't simply vanish between the shells, it becomes constant. Answer B applies the interior-constant rule to the inner shell as well, treating P as if it were inside the inner shell. But P lies outside

R_1

, so you must use

kQ/r

, not

kQ/R_1

. Answer D makes both errors simultaneously — applying the interior formula to the inner shell when P is actually outside it. A useful pattern: always ask "is the field point inside or outside each individual shell?" before writing down its contribution. Inside → use the shell's radius; outside → use the field point's distance.

Question 8

Two point charges $+q$ and $+9q$ are fixed in space separated by a distance $L$ . A third point charge $-q$ is released from rest at the point on the line joining the two positive charges where the electric potential due to the positive charges alone is at its minimum. Which of the following correctly identifies that release point and describes the subsequent motion of $-q$ ?

The minimum potential due to $+q$ and $+9q$ between the charges occurs at $x = L/4$ from the $+q$ charge, which coincides with the point where the net electric field is zero. Since $E = 0$ there, the net force on $-q$ is zero and it remains in unstable equilibrium. (correct answer)
The minimum potential occurs at $x = L/2$ from the $+q$ charge (the geometric midpoint), and the $-q$ charge released there accelerates toward the larger $+9q$ charge because the field from $+9q$ dominates at the midpoint.
The potential due to two positive charges has no minimum between them — it increases monotonically toward each charge and reaches its overall minimum only at infinity. The $-q$ charge would therefore immediately accelerate toward whichever positive charge is closer.
The minimum potential occurs at $x = L/4$ from the $+q$ charge. However, because $-q$ is a negative charge, it moves opposite to the direction a positive test charge would move, so it immediately accelerates away from both positive charges and escapes to infinity.

Explanation: When a question asks about electric potential from multiple charges, your first instinct should be to write the potential explicitly — unlike electric field, potential is a scalar, so you simply add contributions algebraically. The potential due to the two positive charges at a point distance

x

from

+q

(and

L - x

from

+9q

) is:

V(x) = \frac{kq}{x} + \frac{9kq}{L - x}

Both terms are positive and grow without bound as

x \to 0

x \to L

. To find the minimum between the charges, set

dV/dx = 0

-\frac{kq}{x^2} + \frac{9kq}{(L-x)^2} = 0 \implies (L-x)^2 = 9x^2 \implies L - x = 3x \implies x = \frac{L}{4}

So the minimum potential occurs at

x = L/4

from

+q

. This is also precisely where the net electric field from the two positive charges is zero (you can verify by setting the field magnitudes equal). Since

\vec{F} = q\vec{E}

, and

\vec{E} = 0

at that point, the net force on

-q

is zero — making this an equilibrium point. Answer A is correct. Answer B is wrong because it confuses the geometric midpoint with the potential minimum; the stronger charge pulls the minimum closer to the weaker charge, not to the center. Answer C is wrong because the potential does have a minimum between the charges — this reasoning would only apply if one charge were negative, creating a zero in

V

. Answer D identifies the correct location but then incorrectly claims

-q

escapes; since the force is zero at that point, it stays put (in unstable equilibrium). A key study tip: always distinguish between the zero of

V

and the minimum of

V

. For two positive charges,

V > 0

everywhere finite, so there's a minimum between them — not a zero. That minimum coincides with the zero of

E

, which is where force vanishes.

Question 9

Four point charges of equal magnitude $q$ are placed at the corners of a square with side length $L$ . The charges alternate in sign around the square: $+q, -q, +q, -q$ .

What is the electric potential at the exact center of the square?

$V = \frac{4kq}{L}$ , because all four charges are at the same distance $L$ from the center and their magnitudes add directly.
$V = \frac{4\sqrt{2}\,kq}{L}$ , because the distance from each corner to the center is $L\sqrt{2}$ and all four charge magnitudes contribute positively to the scalar sum.
$V = 0$ , because each positive charge at distance $L/\sqrt{2}$ from the center is paired with an equal-magnitude negative charge at the same distance, so the scalar contributions cancel in pairs. (correct answer)
$V = \frac{2kq}{L\sqrt{2}} - \frac{2kq}{L\sqrt{2}}$ , which simplifies to zero only if the charges alternate perfectly; any asymmetry would yield a nonzero result, but under the stated conditions $V \neq 0$ .

Explanation: When a question asks for electric potential at a point due to multiple charges, your first instinct should be to recall that potential is a scalar, not a vector. This distinction is everything here. The electric potential from a point charge is

V = \frac{kq}{r}

, where the sign of

q

matters directly. To find the total potential at any point, you simply add these scalar values — no direction or component analysis needed. For a square with side

L

, the distance from each corner to the center is

\frac{L}{\sqrt{2}}

, which you get from half the diagonal:

\frac{\sqrt{2}L}{2}

. Each of the four charges sits at this same distance. Since the charges alternate

+q, -q, +q, -q

, their contributions are

+\frac{kq}{L/\sqrt{2}}, -\frac{kq}{L/\sqrt{2}}, +\frac{kq}{L/\sqrt{2}}, -\frac{kq}{L/\sqrt{2}}

, which sum to exactly zero. C is correct. Choice A uses the wrong distance (

L

instead of

\frac{L}{\sqrt{2}}

) and ignores the signs of the charges entirely — two separate errors. Choice B correctly identifies the distance but then treats all four charges as positive, forgetting that potential depends on the sign of

q

, not just its magnitude. Choice D actually expresses the right physics — two positive and two negative contributions at equal distances — but then contradicts itself by claiming the result is nonzero. The expression it writes clearly equals zero, making the conclusion wrong. Study tip: Whenever you see "electric potential" (not field), remind yourself: scalars add with sign. Charges of opposite sign at equal distances will always cancel — this is a classic trap disguised as a hard calculation.