Question bank — Define electric field and electric potential

False. Field is the slope of the potential hill, not its height; a hill can be low but steep. Midway between $+Q$ and $-Q$, $V=0$ yet $E$ is large.

False. $V=0$ only says the net accumulated work per charge is zero; the local slope can still be nonzero, so $E\ne 0$ is fully allowed.

False. Inside a hollow charged conductor $E=0$ everywhere yet $V$ is a nonzero constant equal to the surface value — a flat but elevated plateau.

False. Potential is energy per charge, and energy is a plain number (scalar). You add potentials from many charges by simple addition, no arrows.

True. Constant $V$ means no potential difference between start and end, and work per charge equals $\Delta V$; the field is also perpendicular to the path, so it does no work.

False. The whole point of dividing by $q_0$ (the test charge) is that the field is a property of the space, independent of the visitor. Force doubles, but $\vec{E}=\vec{F}/q_0$ stays fixed.

False. $\vec{F}=q\vec{E}$, and a negative $q$ flips the direction — the force points left, opposite the field.

True. $V=kQ/r$ is negative near a negative $Q$ (energy is released, not spent), but field magnitude $E=k|Q|/r^2$ uses $|Q|$ (size only) — a length of an arrow, never negative.

False. A crossing would mean two different force directions at one point, which is impossible — the net field has exactly one direction everywhere it exists.

The formula gives only the magnitude (properly $E=k|Q|/r^2$). Direction is carried by $\hat{r}$: away from positive $Q$, but toward a negative $Q$. Always state direction separately.

You are mixing the two lenses. Potential is a scalar — just add $V_1+V_2$ arithmetically. Vectors and their components belong to the field calculation only.

The force changes at every step (it grows as you approach), so a single $Fr$ is wrong. You must integrate $-{\int }_{\infty }^r{q}_0\vec{E}\cdot d\vec{l}$, which is why $V\propto 1/r$ appears, not $1/r^2$.

Backwards. Potential is the accumulated field over distance ($V=-\int \vec{E}\cdot d\vec{l}$), and integrating $1/r^2$ gives $1/r$ — a slower fall-off, not faster.

Volts are joules per coulomb, not an amount of charge. The 9 V is the potential difference (height of the electricity hill); it sets the energy each coulomb gains, not how much charge moves.

A positive charge rolls from high $V$ to low $V$ (downhill), releasing energy. Only under external work does it climb. A negative charge does the opposite.

The field depends on the derivative (slope) of $V$, not its value. At the very top of a smooth potential peak the slope is zero, so $E=0$ even though $V$ is maximal.

So each quantity describes the space prepared by the source, not the particular visiting charge. The $q_0$ cancels, leaving only $Q$, $r$ and the constant $k$.

Potential encodes energy per charge, which is negative near a negative source (energy released on approach), so we keep $Q$ with its sign. Field magnitude uses $|Q|$ — inherently non-negative; its sign information lives in the direction $\hat{r}$ instead.

It is the fixed proportionality number $k=1/(4\pi {\epsilon }_0)\approx 8.99\times 10^9$ that converts "charge divided by distance" into real newtons and volts; ${\epsilon }_0$ inside it measures how empty space responds to charge. Without $k$ the numbers would have no physical units.

The field points in the direction of decreasing potential (downhill for a positive charge). The minus flips "increasing $V$" into "field pointing the other way," matching the roll-downhill picture.

The single-variable derivative $dV/dr$ measures the slope along $r$ only, so it captures the whole field just when $V$ depends on $r$ alone (spherical symmetry). In general the field is $\vec{E}=-\vec{\nabla }V$, the negative gradient, which stitches together the slope in every direction into one vector.

$d\vec{l}$ is a tiny directed step along the path, and only the part of $\vec{E}$ pointing along that step does work — that projection is the dot product $\vec{E}\cdot d\vec{l}$. For a straight radial pull it collapses to $Edr$, but if $\vec{E}$ ever points sideways or reverses, the dot product automatically tracks the correct sign.

Potential is a scalar (energy per charge), so contributions superpose by ordinary addition. Field is a vector; contributions can point different ways and must be added head-to-tail, respecting direction.

It is not forced — only differences in potential ($\Delta V$) are physically measurable, so we are free to nail the zero anywhere (this is called reference freedom, like choosing sea level for altitude). Infinity is merely the most convenient reference for an isolated charge because $kQ/r\to 0$ there naturally, giving the clean $V=kQ/r$; a lab may instead pick the ground/earth as zero, and every $E$ stays identical because a constant shift in $V$ has zero slope.

$\vec{E}=-\vec{\nabla }V$, so a full $V$-map hands you every slope by differentiation. A single $E$ value is one slope, but potential is an accumulated integral — you need the field along an entire path to rebuild it.

The two fields are equal and opposite, so $E=0$. But the two potentials are both positive and add, so $V=2kQ/d_{half}\ne 0$ — a case of zero field with nonzero potential.

Potentials cancel ($+$ and $-$ add to $V=0$), yet both field arrows point the same way (toward the $-Q$), so they add to a large nonzero $E$ — the mirror-image trap.

Both diverge — $V\to \infty$ as $1/r$ and $E\to \infty$ as $1/r^2$. This is a sign the ideal point charge is an idealisation; real charges have finite size, so nothing physically reaches infinity.

$E=0$ throughout the cavity (fields cancel), but $V$ is a nonzero constant equal to the surface potential — a flat, elevated plateau where the slope is zero but the height is not.

Since $E$ is constant and $E_r=-dV/dr$, the potential changes linearly (constant slope) from one plate to the other — steady, not $1/r$, because this is not a point-charge geometry.

Both approach zero — $V\to 0$ (our chosen reference) and $E\to 0$. Far away the space is essentially "unprepared," no push and no stored energy per charge.

Nothing accelerates it, because constant $V$ means zero slope means $E=0$, so the force $q\vec{E}=0$. It stays put — a flat spot on the hill.