Question bank — Electric field — definition, field lines, superposition

False. It accelerates along $\vec{E}$, but it also has inertia and prior velocity — its path follows a line only if it starts from rest; otherwise it drifts across lines like a ball thrown sideways in gravity.

True. By definition $\vec{E}$ is force-per-positive-charge, and the tangent to a line is $\vec{E}$, so a positive charge is pushed along the line's arrow.

False. $\vec{E}$ can vanish at a point in empty space — e.g. midway between two equal positive charges, where the two fields cancel. Zero field ≠ zero nearby charge.

False. $\vec{E}=\vec{F}/q_0$: the force doubles too, so the ratio is unchanged. That is the whole point of dividing out $q_0$ — the field belongs to the sources, not the probe.

False. Touching means sharing a point with the same tangent direction; that's still $\vec{E}$ having one value there, which is fine — but they then coincide, they don't remain two distinct lines. Genuine crossing (two directions at one point) is forbidden.

True. $\vec{E}=kq/r^2$ blows up as $r\to 0$; the formula gives infinity, so the point charge's own location is a singularity (an idealization real charge spreads out avoids).

False. Lines enter (point toward) negative charges. A positive test charge near a negative source is pulled inward, so $\vec{E}$ — and the line's arrow — points toward the negative charge.

False. The total number you draw is only a convention; what encodes strength is line density (lines per unit area crossed). Doubling the charge doubles lines and can spread them thinner, so a bigger count alone does not guarantee a stronger field.

Error: those are vectors. You must resolve each into $x,y$ (and $z$) components and add componentwise. Magnitudes add only if all three fields happen to be parallel.

Error: $\hat{r}$ points from source to field point. Then the sign of $q$ handles direction: positive $q$ gives $\vec{E}$ along $\hat{r}$ (outward), negative $q$ flips it inward.

Error: a charge exerts no net force on itself. $\vec{E}$ at a point is built only from the other (source) charges evaluated at that location.

Error: it's the reverse — set axes with the charges on the x-axis (see figure s01, right panel). The two y-components have opposite signs and cancel; the two x-components both point in $-x$ and add. See Electric Dipole.

Error: only if the net charge is nonzero. A dipole (zero net charge) falls as $1/r^3$; a quadrupole faster still. The leading term is the lowest nonzero multipole.

Error: the field is defined by the push on a positive test charge, which is repelled — so $\vec{E}$ of a positive source points outward. Attraction of opposites comes from the negative test charge, not the source's field.

Error: you add the vector fields ${\vec{E}}_i$, each already computed with its own $r_i$. You never combine denominators — every source keeps its own distance and direction ${\hat{r}}_i$. This is Coulomb's Law applied once per source, then summed.

So the field describes only the source and location, independent of what probe we use. Dividing removes the probe from the answer, making $\vec{E}$ a property of space itself.

A real test charge pushes back on the sources and relocates them, altering the very field we want. The limit ensures the probe reads without disturbing.

The same fixed number of lines from a point charge spreads over a sphere of area $4\pi r^2$. Density = lines/area $\propto 1/r^2$ — the geometry is the $1/r^2$ falloff of Coulomb's Law.

Because Coulomb's law is linear in charge and forces add as vectors (Newton). Linearity means each source acts as if the others weren't there, and results just sum — the essence of the Superposition Principle.

Far away the $+q$ and $-q$ nearly cancel; only the tiny leftover from their separation survives, and that residue shrinks one power of $r$ faster. See Electric Dipole.

Instead of charge B mysteriously "knowing" about distant charge A, A fills space with a field and B responds only to the field right where B sits — no action across a gap.

The lines are a sampling of a continuous vector field. Between any two lines the field still exists; we draw a finite few for readability, choosing density to encode strength.

Zero. The two fields are equal in magnitude and opposite in direction (figure s01, left), so they cancel — a positive test charge there feels no push (an unstable equilibrium).

No. Both fields point the same way (from $+$ toward $-$, figure s01 right) and add. Only when charges have the same sign do the midpoint fields oppose and cancel.

It tends to zero like $1/r^2$, but never exactly reaches zero — the field has infinite range, just vanishingly weak far away.

Yes — at that point the two source contributions superpose into one resultant vector with a single direction. The apparent "meeting" is just where one net line passes; the field is perfectly single-valued.

None ($\vec{E}=0$ everywhere from it). With no charge there is no source term in $kq/r^2$; it neither emits nor terminates field lines.

No — the field is just weak, not absent. Low line density signals small $|\vec{E}|$; a line still passes through every point in principle.

Zero everywhere inside. Charges rearrange on the conductor's surface until the interior field cancels completely — a classic result of electrostatic equilibrium (provable with Gauss's Law). This is why a metal box shields its interior.

Perpendicular to the surface. Any tangential component would push the surface charges sideways until they stop moving, so at equilibrium only the perpendicular part survives.