Coulomb's law — force, comparison with gravity: Question bank

Notation and pictures you'll need first

Several traps below hinge on notation the parent used quickly. Let's pin it down with pictures so no symbol is a mystery when it appears.

Figure — Coulomb's law — force, comparison with gravity

Look at the figure: the violet arrow is $r_{21}$ running from 1 to 2, and the short magenta arrow is $\overset{r}{^}_{21}$ — same direction, length 1. Everything about "which way does the force point" is decided by this little arrow.

True or false — justify

Doubling the separation

r

halves the Coulomb force.

False. The law goes as

1/ r^{2}

, not

1/ r

; doubling

r

multiplies force by

1/ 2^{2} = 1/4

. "Twice as far → half as strong" is linear intuition, but electric influence spreads over a sphere of area

4 π r^{2}

, so it thins as the square.

If both charges are negative, the force between them is negative (attractive).

False. Two negatives repel. In the scalar form we use

∣ q_{1} q_{2} ∣

, so the magnitude is positive; the product

(-) (-) = +

signals like charges → repulsion. A "negative force" is not a thing on its own — direction lives in the vector form, not the sign of a magnitude.

The

r^{2}

F_{e} / F_{g}

for two electrons cancels only if they are far apart.

False. It cancels always, at every distance. Both

F_{e}

and

F_{g}

carry the identical

1/ r^{2}

, so their ratio is a pure constant (

\approx 4.2 \times 1 0^{42}

) independent of

r

Coulomb's law applies exactly to two charged metal spheres regardless of size.

False in general. It is exact for point charges, or for spherically symmetric charge treated as if concentrated at the centre. Two large spheres held close redistribute each other's charge (induction), breaking the symmetry, so the simple formula becomes approximate.

Gravity is stronger than the electric force whenever objects are large.

Misleading. Gravity dominates at large scale not because it grows stronger, but because big bodies are electrically neutral (equal

+

and

-

), so their vastly stronger electric forces cancel, leaving feeble gravity uncancelled.

The constant

k

being large (

8.99 \times 1 0^{9} N\cdotm^{2} / C^{2}

) and

G

being tiny (

6.67 \times 1 0^{- 11} N\cdotm^{2} / kg^{2}

) alone proves electricity beats gravity.

False as stated. The two constants carry different units (per C² vs per kg²), so you cannot compare the bare numbers directly. The honest comparison plugs in real charges and masses of the same particles — only then does electricity win by

\sim 1 0^{42}

Newton's third law fails for Coulomb forces because one charge might be bigger.

False.

F_{12} = - F_{21}

always: the force on 1 from 2 equals in size and opposes in direction the force on 2 from 1, even if

q_{1} \neq = q_{2}

. Both share the same

∣ q_{1} q_{2} ∣

and the same

r

A charge feels its own electric force.

False. Coulomb's law is about the force between two charges. A charge does not push on itself; you always need a second (source) charge.

Spot the error

F = k \frac{q _{1} q _{2}}{r ^{2}}

gave me

- 5.4 N

, so the force is

- 5.4

newtons."

The scalar magnitude formula must use

∣ q_{1} q_{2} ∣

; a magnitude can't be negative. The minus sign told you the charges are unlike (attractive), but that is a direction conclusion, read off separately — not a negative newton value.

"The charges are

3 μ C

and

2 μ C

, so I plug in

3

and

2

Units error. Coulomb's law needs SI:

3 μ C = 3 \times 1 0^{- 6} C

. Forgetting the

1 0^{- 6}

inflates the answer by

1 0^{12}

"They're

10 cm

apart, so

r^{2} = 1 0^{2} = 100

Wrong units for

r

. Convert first:

10 cm = 0.1 m

, so

r^{2} = 0.01 m^{2}

. Squaring in centimetres corrupts everything downstream.

\overset{r}{^}_{21}

points from 2 to 1, since it's on charge 2."

Reversed. As defined above,

r_{21} = r_{2} - r_{1}

points from 1 to 2 (source → target); the subscript order

21

reads "on 2 due to 1." Getting this backwards flips every attract/repel conclusion.

"Halving both charges and halving the distance leaves the force unchanged."

The arithmetic is right but there is no rule to remember here. It happened that two separate effects — charges

\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}

and distance

1/ (\frac{1}{2})^{2} = \times 4

— multiplied to exactly

1

for these particular factors. This is a lucky coincidence of the numbers chosen, not a principle: pick any other factors (say quarter the charge, halve the distance) and the force changes. Never memorise this as "it cancels."

"Coulomb's force needs the charges to be in contact to act."

The whole point is that it is a force at a distance — no contact required. The charges "feel" each other across empty space (via the electric field, the invisible arrow-map each charge sets up around itself).

"With three charges, I just add the two forces' magnitudes."

Force is a vector. With more than one source you must add forces as vectors — the Superposition Principle — respecting direction (see the figure below); scalar addition only works if all forces happen to point the same way.

In the figure, a test charge feels force $F_{A}$ from source A and $F_{B}$ from source B. The correct total is the tip-to-tail resultant $F_{A} + F_{B}$ (orange), which is shorter than $∣ F_{A} ∣ + ∣ F_{B} ∣$ because the two arrows point in different directions. Adding magnitudes would overcount.

Why questions

Why does Coulomb's law contain

1/ r^{2}

rather than

1/ r

1/ r^{3}

The charge's influence spreads uniformly over the surface of an expanding sphere whose area is

4 π r^{2}

(see the sphere figure below). Conserving total influence spreads it thinner as area grows, giving exactly

1/ r^{2}

— the geometry of 3D space picks the exponent.

The figure shows the same fixed number of "influence lines" leaving a charge and piercing two spheres. On the bigger sphere the same lines are spread over area $4 π r^{2}$ , so their density — and hence the force — falls as $1/ r^{2}$ . This picture is exactly the seed of Gauss's Law, which counts those lines through any closed surface.

Why is

k

written as

\frac{1}{4 π ε _{0}}

instead of just a single number?

The

4 π

is the surface area factor of a unit sphere (the same

4 π r^{2}

from the spreading picture). Building it into

k

makes later laws — especially Gauss's Law, which sums field-line "flux" over a closed surface — come out clean and factor-free, worth the messier-looking constant here.

Why can the electric force both push and pull while gravity only pulls?

Charge comes in two signs (

+

and

-

); like signs repel, unlike attract. Mass comes in only one sign (there is no negative mass), so gravity can only ever attract.

Why do we use signed charges in the vector form but magnitudes in the scalar form?

In the vector form

F_{21} = k \frac{q _{1} q _{2}}{r ^{2}} \overset{r}{^}_{21}

, the sign of

q_{1} q_{2}

automatically flips the arrow (along or against

\overset{r}{^}_{21}

) to give repel vs attract. The scalar form only reports size, so it takes

∣ q_{1} q_{2} ∣

and you decide direction by hand.

Why is the electric force between two protons in a nucleus a "problem" nature had to solve?

At nuclear separations the Coulomb repulsion between protons is enormous (hundreds of newtons on a subatomic speck). Something stronger — the strong nuclear force — must overcome it to hold the nucleus together.

Why doesn't the huge electric force between everyday objects tear them apart?

Ordinary matter is electrically neutral: every atom's positive nucleus is balanced by its electrons. The colossal forces are present internally but cancel almost perfectly, leaving only tiny residual effects.

Why does distance "not matter" when comparing electric to gravitational force for the same pair?

Both forces scale as

1/ r^{2}

identically, so when you divide one by the other the

r^{2}

terms cancel exactly, leaving a distance-independent ratio.

Edge cases

What is the Coulomb force if one of the charges is exactly zero?

Zero.

F \propto q_{1} q_{2}

, so a neutral (uncharged) object experiences no Coulomb force from a charge — there is nothing for the field to act on.

What happens to the predicted force as

r \to 0

(charges brought together)?

The formula gives

F \to \infty

, because

1/ r^{2}

blows up. Physically this signals the breakdown of the point-charge idealisation: real charges have size and other forces intervene before infinity is reached.

What is the force as

r \to \infty

(charges very far apart)?

It tends to zero but never reaches it.

1/ r^{2}

shrinks fast, so the force becomes negligible at large distance — yet formally it is nonzero at every finite separation (infinite range).

If two charges are equal and opposite (like

+ q

and

- q

), is the force on each different?

No. By Newton's third law the two forces are equal in magnitude and opposite in direction. Both share

∣ q_{1} q_{2} ∣ = q^{2}

and the same

r

, so neither "wins."

Does making both charges negative change the magnitude of the force compared to both positive (same sizes)?

No. Magnitude depends only on

∣ q_{1} ∣∣ q_{2} ∣

, so

(+ q, + q)

and

(- q, - q)

give identical magnitudes — both repulsive. Only the sign pattern (like vs unlike), not which sign specifically, sets attraction vs repulsion.

Can Coulomb's law be "shielded" the way gravity cannot?

Yes. Surrounding a region with conductors or arranging opposite charges can cancel or block an electric field, so electric forces are shieldable — there is no known way to shield gravity, since mass has only one sign.

If you double one charge but flip its sign, what happens to the force?

Its magnitude doubles (since

∣ q ∣

doubled) and its direction reverses (like→unlike or vice versa). Both effects happen at once: stronger and the opposite kind (attract↔repel).

Recall One-line summary of the traps

Squared not linear distance ::: force $\propto 1/ r^{2}$ , so double $r$ → quarter force. Magnitudes vs signs ::: $∣ q_{1} q_{2} ∣$ for size, sign of product for attract/repel. Point charges only ::: exact for points or spherical symmetry, else integrate. Neutrality, not weakness ::: gravity dominates cosmically because charges cancel, not because electricity is weak. $k$ vs $G$ ::: different units (per C² vs per kg²), so bare numbers don't compare.