Gravitational vs Electric Force: The Difference

Both forces fall off as one over the distance squared and act along the line joining the two objects. Gravity is always attractive and depends on mass; the electric force attracts or repels and depends on charge. It is also vastly stronger, and charge can be cancelled while mass cannot.

AP Physics: Unit 10 (topics 10.1 Electric Charge and Electric Force, 8.1 Electric Charge and Electric Force, 2.6 Gravitational Force). The comparison is required content in AP Physics 2 and AP Physics C: Electricity and Magnetism, and in both it sits in the first topic of the first electricity unit. AP Physics 2 Unit 10, Electric Force, Field, and Potential, is weighted at 15 to 18 percent of the multiple-choice section over about 14 to 21 class periods; Topic 10.1 carries learning objective 10.1.B, describe the electric and gravitational forces that result from interactions between charged objects with mass, with EK 10.1.B.1 that electrostatic forces can be attractive or repulsive while gravitational forces are always attractive, EK 10.1.B.2 that for any two objects that have mass and electric charge the magnitude of the gravitational force is usually much smaller than the magnitude of the electrostatic force, and EK 10.1.B.3 that gravitational forces dominate at larger scales even though they are weaker, because systems at large scales tend to be electrically neutral. Coulomb's law is EK 10.1.A.2 with the relevant equation printed on the sheet, and EK 10.1.A.3 with its two sub-statements sets the direction from the signs. Suggested skills for Topic 10.1 are 1.A, 2.A, 2.D and 3.B. AP Physics C: Electricity and Magnetism Unit 8, weighted at 15 to 25 percent over about 12 to 24 class periods, repeats all of this at Topic 8.1 with LO 8.1.B and EK 8.1.B.1 through 8.1.B.3, with suggested skills 1.A, 2.B, 2.D and 3.B. The Topic 10.1 boundary statement limits AP Physics 2 to calculations of the electric force between four or fewer interacting charged objects or systems, with more charges allowed in situations of high symmetry. The gravitational half comes from EK 2.6.A.1 in AP Physics 1 and C: Mechanics. Coulomb's law and k = 9.0 times 10 to the ninth are printed only on the Physics 2 and C: E and M sheets; the gravitational force law and G are printed on all four.

The CED asks you to compare them, and it names three differences

This comparison is not something a textbook invented. AP Physics 2 learning objective 10.1.B reads: describe the electric and gravitational forces that result from interactions between charged objects with mass. AP Physics C: Electricity and Magnetism carries the identical objective at 8.1.B, with the same three essential knowledge statements word for word.

Here they are, because they are the answer to the page's question and the CED states them more tightly than most summaries manage:

EK 10.1.B.1 / 8.1.B.1: electrostatic forces can be attractive or repulsive, while gravitational forces are always attractive.
EK 10.1.B.2 / 8.1.B.2: for any two objects that have mass and electric charge, the magnitude of the gravitational force is usually much smaller than the magnitude of the electrostatic force.
EK 10.1.B.3 / 8.1.B.3: gravitational forces dominate at larger scales even though they are weaker than electrostatic forces, because systems at large scales tend to be electrically neutral.

Sign, strength, and why the weaker one runs the universe. Notice that the third statement is not a fourth independent difference but the resolution of the tension between the first two: the electric force wins every head-to-head contest and loses every large-scale one, because the only force that cannot be cancelled is the one that only ever pulls.

The resemblance that makes the comparison worth drawing is on the sheets. Both force laws are inverse-square laws of the same shape, with a coupling constant, a product of source properties, and $r^2$ underneath:

\lvert \vec{F}_g \rvert = G\frac{m_1 m_2}{r^2} \qquad\qquad \lvert \vec{F}_E \rvert = \frac{1}{4\pi\varepsilon_0}\frac{\lvert q_1 q_2 \rvert}{r^2} = k\frac{\lvert q_1 q_2 \rvert}{r^2}

The Coulomb's law guide carries the routine for working the electric one. This page is about what the two laws do differently once you have both.

Gravitational vs electric force, side by side

Question you are asking	Gravitational force	Electric force
CED essential knowledge	2.6.A.1	10.1.A.2 and 8.1.A.2
Equation	$\lvert \vec{F}_g \rvert = G\dfrac{m_1 m_2}{r^2}$	$\lvert \vec{F}_E \rvert = k\dfrac{\lvert q_1 q_2 \rvert}{r^2}$
Source property	Mass, in kg	Charge, in C
Signs the source property can take	One, always positive	Two, positive and negative
Coupling constant	$G = 6.67 \times 10^{-11}$	$k = 9.0 \times 10^9$
Direction	Always attractive, EK 2.6.A.1.i	Attractive or repulsive by sign, EK 10.1.A.3
Acts along	The line joining the centers of mass	The line of separation, EK 10.1.A.3
Distance dependence	$1/r^2$	$1/r^2$
Obeys Newton's third law	Yes	Yes
Superposes as a vector sum	Yes	Yes
Can be cancelled to zero by the source	No	Yes, by neutrality
Can be screened by matter in between	No	Yes, EK 10.3.B.1
Depends on the surrounding medium	No	Yes, through permittivity, EK 10.1.C
Its field	$\vec{g} = \vec{F}_g / m$	$\vec{E} = \vec{F}_E / q$
Printed on which sheets	All four booklets	AP Physics 2 and C: E and M only
Dominates at	Astronomical scales	Atomic and everyday contact scales

Four rows carry the page.

The signs row is the structural difference, and it is not about arithmetic. Mass has one sign, so gravitational contributions from a large body all add. Charge has two, so contributions from a large body mostly cancel. That single asymmetry is why the ratio of coupling constants, twenty orders of magnitude, does not decide which force shapes the solar system.

The screening row has no gravitational counterpart at all. EK 10.3.B.1 says that while in electrostatic equilibrium the excess charge of a solid conductor is distributed on the surface of the conductor, and the electric field within the conductor is zero. Put a sensitive charge inside a metal box and the outside world's electric field does not reach it. There is no material that does this for gravity, and none is expected to exist.

The medium row is a difference students rarely meet as one. EK 10.1.C.1 defines electric permittivity as a measurement of the degree to which a material or medium is polarized in the presence of an electric field, EK 10.1.C.3 says free space has a constant value $\varepsilon_0$ , and EK 10.1.C.4 says the permittivity of matter differs from that of free space, arising from the matter's composition and arrangement. So the electric force between two charges genuinely depends on what is between them. $G$ has no such companion.

The sheets row is worth checking rather than assuming. The gravitational force law is printed in the mechanics table of all four booklets, including the AP Physics C: Electricity and Magnetism booklet, which carries a full mechanics table. Coulomb's law is printed only on the AP Physics 2 and AP Physics C: Electricity and Magnetism sheets, and so is $k = 9.0 \times 10^9$ . The AP Physics 1 and C: Mechanics constants boxes have $G$ and no Coulomb constant.

The strength ratio is a pure number, and it is different for every pair

Here is the fact that makes EK 10.1.B.2 precise. Take any two objects with mass and charge and form the ratio of the two forces between them:

\frac{F_E}{F_g} = \frac{k \lvert q_1 q_2 \rvert / r^2}{G m_1 m_2 / r^2} = \frac{k}{G} \cdot \frac{\lvert q_1 q_2 \rvert}{m_1 m_2}

The $r^2$ cancels. Both laws are inverse square with the same exponent, so the ratio does not depend on how far apart the objects are. It is a pure number fixed by the two objects, and it is the same at a nanometre as at a light year. That is why the statement holds at every scale and why nothing about distance can rescue gravity in a two-body contest.

What the number is depends entirely on the charge-to-mass ratios, and those vary enormously. Using the constants box values $k = 9.0 \times 10^9$ , $G = 6.67 \times 10^{-11}$ , $e = 1.60 \times 10^{-19}$ C, $m_p = 1.67 \times 10^{-27}$ kg and $m_e = 9.11 \times 10^{-31}$ kg, all printed on the AP Physics 2 sheet:

Pair	$F_E / F_g$
Two protons	$1.2 \times 10^{36}$
Two electrons	$4.2 \times 10^{42}$

Worked example one computes both digit by digit. The gap between the two rows is worth staring at: the charges are identical in magnitude, so the entire six-order-of-magnitude difference comes from the masses, and it is exactly $(m_p/m_e)^2 = 3.4 \times 10^6$ .

So there is no single number that answers how much stronger the electric force is. Anyone who quotes one has picked a pair of particles. The structural fact is the one worth carrying: the ratio is $\dfrac{k}{G}$ , which is about $1.3 \times 10^{20}$ , multiplied by the ratio of the charge product to the mass product.

That second factor is what neutrality attacks. Worked example two takes two ordinary $1.0$ kg spheres and asks how much charge each would need for the two forces to be equal, and the answer is $8.6 \times 10^{-11}$ C, about $5.4 \times 10^8$ electrons. Those electrons together weigh $4.9 \times 10^{-22}$ kg, so removing them changes each sphere's mass by less than one part in $10^{21}$ . A charge imbalance far too small to weigh is enough to make the electric force match gravity. Ordinary matter is neutral to better than that, which is the content of EK 10.1.B.3.

The case that separates them: what happens when you change the sign

Two identical small spheres, each of mass $0.15$ kg, sit $0.30$ m apart and each carries a charge of $+2.0$ nC.

Force	Magnitude	Direction
Gravitational	$1.67 \times 10^{-11}$ N	Attractive, toward each other
Electric	$4.0 \times 10^{-7}$ N	Repulsive, apart
Net	$4.0 \times 10^{-7}$ N	Repulsive

The electric force is about $2.4 \times 10^4$ times the gravitational one, so the net is repulsive and the gravitational contribution does not show up in the first three significant figures.

Now flip the sign of one charge and change nothing else.

Force	Magnitude	Direction
Gravitational	$1.67 \times 10^{-11}$ N	Attractive, unchanged
Electric	$4.0 \times 10^{-7}$ N	Attractive
Net	$4.0 \times 10^{-7}$ N	Attractive

One row changed and one row did not, and that is the experiment. The gravitational force did not notice, because there is no operation you can perform on a mass that reverses the direction of gravity. EK 2.6.A.1.i is a single flat sentence: the gravitational force is attractive. The electric force reversed completely, because EK 10.1.A.3 makes its direction depend on the signs of the charges, with 10.1.A.3.i giving repulsion for like signs and 10.1.A.3.ii giving attraction for opposite signs.

The third configuration is the one with no gravitational analogue at all. Neutralise the spheres and the electric force goes to exactly zero while the gravitational force is untouched. You can turn one of these two forces off and leave the objects in place. You cannot do that to the other one.

Worked example three runs all three configurations and also finds the charge at which the two forces balance for these particular spheres: $1.29 \times 10^{-11}$ C, about $155$ times smaller than the $2.0$ nC they started with. That is the sense in which EK 10.1.B.2's word usually is doing careful work. The gravitational force is smaller for any pair of objects carrying charge worth mentioning, and it wins only when the charge is essentially absent.

Why the weaker force runs the universe

EK 10.1.B.3 gives the reason in one clause: systems at large scales tend to be electrically neutral. It is worth unpacking into the three mechanisms that keep them that way, because each one is a piece of course content elsewhere.

Charge comes in two signs and they cancel. A kilogram of matter contains an enormous quantity of positive and negative charge in almost exact balance. Doubling the amount of matter doubles both, so the net stays near zero while the mass grows without limit. Mass has no negative version to cancel against, so the gravitational contribution of every gram simply adds.

Excess charge does not stay put. EK 10.2.A.3 says grounding involves electrically connecting a charged system to a much larger and approximately neutral system, and the CED's example of the larger system is Earth. Any large charged body sits in a universe full of mobile charge that will flow to neutralise it. Nothing analogous drains mass away from a massive body.

Conductors screen their interiors. EK 10.3.B.1 states that while in electrostatic equilibrium the excess charge of a solid conductor is distributed on the surface of the conductor, and the electric field within the conductor is zero, with EK 10.3.B.1.i adding that at the surface of a charged conductor the electric field is perpendicular to the surface. So even where charge is present, matter can hide from it.

The consequence is a clean division of labour that the CED sets up in EK 10.1.A.4, and this statement is worth reading closely because it reframes most of the forces in AP Physics 1. Electric forces are responsible for some of the macroscopic properties of objects in everyday experiences. However, the large number of particle interactions that occur make it more convenient to treat everyday forces in terms of nonfundamental forces called contact forces, such as normal force, friction, and tension.

Read against this page's comparison, that says something surprising. The normal force holding you up, the friction stopping your shoe, the tension in a rope: all of them are the electric force, packaged. So the electric force is not confined to the electricity units. It is doing almost all of the pushing in the mechanics units too, under other names. Gravity, meanwhile, is the only one of the two that ever appears undisguised in a mechanics problem, and it is the one that shapes orbits, because at planetary distances the electric contribution has cancelled itself to nothing.

Where the confusion costs a mark

Each of these is a specific scoring error.

Putting a sign into Coulomb's law and reading the answer's sign as a direction. The printed form uses the absolute value of the charge product, $\lvert q_1 q_2 \rvert$ , so it returns a magnitude. The direction comes from EK 10.1.A.3 and the signs of the charges, decided separately and stated in words or on a diagram.
Writing the gravitational force as negative to mean attractive. $\lvert \vec{F}_g \rvert = Gm_1m_2/r^2$ is a magnitude too. Attraction is a direction, not a minus sign, and the minus sign in $U_G = -Gm_1m_2/r$ is a different statement about potential energy.
Using $k$ where $G$ belongs, or the reverse. They differ by twenty orders of magnitude, so the arithmetic error is not subtle, but it is easy to make under time pressure when the two formulas look alike.
Using $8.99 \times 10^9$ for the Coulomb constant. The AP sheets print $k = 1/(4\pi\varepsilon_0) = 9.0 \times 10^9\ \text{N}\cdot\text{m}^2/\text{C}^2$ . Use the printed value.
Quoting one universal number for how much stronger electricity is. There is no such number. The ratio depends on the two objects' charge-to-mass ratios, and it differs by a factor of $3.4 \times 10^6$ between a pair of protons and a pair of electrons.
Saying the ratio depends on separation. It does not. Both laws carry the same $r^2$ , so it cancels out of the ratio entirely.
Dropping the gravitational force from a charged-particle problem without saying so. Neglecting it is right and it is worth one sentence of justification, which is exactly what EK 10.1.B.2 supplies.
Dropping the electric force from a mechanics problem while using the normal force. By EK 10.1.A.4 the contact forces are the electric force, repackaged, and a free-response answer that claims electricity is absent from a collision problem has said something false.
Saying gravity is stronger because planets orbit. EK 10.1.B.3 says the opposite in the same sentence as the observation: gravitational forces dominate at larger scales even though they are weaker than electrostatic forces, because systems at large scales tend to be electrically neutral. Dominant and stronger are different claims.
Measuring $r$ from a surface. For gravity, EK 2.6.A.1.ii puts the line between centers of mass. For the electric force, the point-charge model at EK 10.1.A.1.iv treats the object's physical size as negligible in the first place.
Exceeding the exam's scope. The Topic 10.1 boundary statement says AP Physics 2 only expects students to make calculations of the electric force between four or fewer interacting charged objects or systems, with the analysis of the resulting electric force from more charges allowed in situations of high symmetry.

Where each course asks for this comparison

The comparison is required content in two of the four courses, and both of them state it in the same place: the first topic of their first electricity unit.

AP Physics 2 Unit 10, Electric Force, Field, and Potential: 15 to 18 percent of the multiple-choice section, about 14 to 21 class periods. Topic 10.1, Electric Charge and Electric Force, carries LO 10.1.A on the electric force itself, LO 10.1.B, describe the electric and gravitational forces that result from interactions between charged objects with mass, and LO 10.1.C on electric permittivity. Suggested skills for Topic 10.1 are 1.A, 2.A, 2.D and 3.B.

AP Physics C: Electricity and Magnetism Unit 8, Electric Charges, Fields, and Gauss's Law: 15 to 25 percent, about 12 to 24 class periods. Topic 8.1, Electric Charge and Electric Force, carries LO 8.1.A, LO 8.1.B and LO 8.1.C with the same essential knowledge statements as Physics 2's Topic 10.1. Suggested skills for Topic 8.1 are 1.A, 2.B, 2.D and 3.B.

AP Physics 1 and AP Physics C: Mechanics do not cover the electric force, so the comparison does not appear in their frameworks. Their booklets still print the gravitational force law and $G$ , and their Topic 2.6 supplies the gravitational half of everything on this page.

On the equation sheets, checked against the appendix pages of all four Course and Exam Descriptions:

Booklet	$\lvert \vec{F}_g \rvert = Gm_1m_2/r^2$	Coulomb's law	$G$ in the constants box	$k$ in the constants box
AP Physics 1	Printed	Not printed	Yes	No
AP Physics 2	Printed	Printed	Yes	Yes
AP Physics C: Mechanics	Printed	Not printed	Yes	No
AP Physics C: E and M	Printed	Printed	Yes	Yes

The field versions of the two laws sit one line apart on the Physics 2 and C: E and M sheets in the same pattern, $\vec{E} = \vec{F}_E/q$ against the gravitational $\lvert \vec{g} \rvert = \lvert \vec{F}_g \rvert/m$ , which is the subject of gravitational field vs gravitational force. For the CED framing see Topic 10.1 in AP Physics 2, Topic 8.1 in C: Electricity and Magnetism and Topic 2.6 in AP Physics 1. The Coulomb's law calculator will do the arithmetic.

The force ratio for two protons and for two electrons

Find the ratio of the electric force to the gravitational force between two protons, and then between two electrons, at any separation. Use the AP constants box values $k = 9.0 \times 10^9\ \text{N}\cdot\text{m}^2/\text{C}^2$ , $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ , $e = 1.60 \times 10^{-19}$ C, $m_p = 1.67 \times 10^{-27}$ kg and $m_e = 9.11 \times 10^{-31}$ kg. Then show why the answer does not depend on the separation.

Set up the ratio symbolically first, because the separation is about to disappear. $\dfrac{F_E}{F_g} = \dfrac{k q^2 / r^2}{G m^2 / r^2} = \dfrac{k q^2}{G m^2}$ . Both laws carry $1/r^2$ , so the $r^2$ divides out exactly, and the ratio is a property of the pair of particles alone.
Electric numerator, shared by both cases. $e^2 = (1.60 \times 10^{-19})^2 = 2.56 \times 10^{-38}\ \text{C}^2$ . Then $k e^2 = (9.0 \times 10^9)(2.56 \times 10^{-38}) = 2.304 \times 10^{-28}\ \text{N}\cdot\text{m}^2$ .
Gravitational denominator for two protons. $m_p^2 = (1.67 \times 10^{-27})^2 = 2.7889 \times 10^{-54}\ \text{kg}^2$ . Then $G m_p^2 = (6.67 \times 10^{-11})(2.7889 \times 10^{-54}) = 1.860 \times 10^{-64}\ \text{N}\cdot\text{m}^2$ .
Divide. $\dfrac{F_E}{F_g} = \dfrac{2.304 \times 10^{-28}}{1.860 \times 10^{-64}} = 1.24 \times 10^{36}$ , that is about $1.2 \times 10^{36}$ .
Gravitational denominator for two electrons. $m_e^2 = (9.11 \times 10^{-31})^2 = 8.299 \times 10^{-61}\ \text{kg}^2$ . Then $G m_e^2 = (6.67 \times 10^{-11})(8.299 \times 10^{-61}) = 5.536 \times 10^{-71}\ \text{N}\cdot\text{m}^2$ .
Divide. $\dfrac{F_E}{F_g} = \dfrac{2.304 \times 10^{-28}}{5.536 \times 10^{-71}} = 4.16 \times 10^{42}$ , that is about $4.2 \times 10^{42}$ .
Check the gap between the two answers. The charges were identical, so the whole difference must come from the masses: $\dfrac{4.16 \times 10^{42}}{1.24 \times 10^{36}} = 3.36 \times 10^{6}$ , and $\left(\dfrac{m_p}{m_e}\right)^2 = \left(\dfrac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}}\right)^2 = (1833)^2 = 3.36 \times 10^6$ . They agree, which confirms the ratio depends on charge and mass only.
Put an absolute number on it once, to see that both forces are real. At a separation of $1.0 \times 10^{-15}$ m, the electric repulsion between two protons is $F_E = (2.304 \times 10^{-28})/(1.0 \times 10^{-30}) = 230\ \text{N}$ , while the gravitational attraction is $F_g = (1.860 \times 10^{-64})/(1.0 \times 10^{-30}) = 1.9 \times 10^{-34}\ \text{N}$ . A force of $230$ N between two protons is the everyday scale of a person's weight, and it is why nuclei need something stronger still to hold together.

For two protons $F_E/F_g = 1.2 \times 10^{36}$ ; for two electrons $F_E/F_g = 4.2 \times 10^{42}$ . Both ratios are independent of separation, because the $r^2$ in the two inverse-square laws cancels. The factor of $3.4 \times 10^6$ between the two is exactly the square of the proton-to-electron mass ratio, so there is no single number that says how much stronger the electric force is.

How little charge it takes to match gravity

Two spheres, each of mass $1.0$ kg, sit with their centers $1.0$ m apart. Find the gravitational force between them. Then find the electric force if each carries a charge of $1.0\ \mu\text{C}$ . Finally, find the charge each would need for the two forces to be equal in magnitude, express it as a number of electrons, and find the mass of those electrons.

Gravitational force. $\lvert \vec{F}_g \rvert = G\dfrac{m_1m_2}{r^2} = \dfrac{(6.67 \times 10^{-11})(1.0)(1.0)}{(1.0)^2} = 6.67 \times 10^{-11}\ \text{N}$ . That is about $67$ piconewtons, at the very edge of what a laboratory torsion balance can measure.
Electric force with $1.0\ \mu\text{C}$ each. $\lvert \vec{F}_E \rvert = k\dfrac{\lvert q_1 q_2 \rvert}{r^2} = \dfrac{(9.0 \times 10^9)(1.0 \times 10^{-6})^2}{(1.0)^2} = (9.0 \times 10^9)(1.0 \times 10^{-12}) = 9.0 \times 10^{-3}\ \text{N}$ .
Compare. $\dfrac{9.0 \times 10^{-3}}{6.67 \times 10^{-11}} = 1.35 \times 10^{8}$ . A microcoulomb, a charge small enough that nothing about the spheres looks different, produces a force $135$ million times the gravitational one.
Now solve for equality. Setting $k q^2 / r^2 = G m^2 / r^2$ cancels the $r^2$ again and gives $q = m\sqrt{G/k}$ . The square root is $\sqrt{(6.67 \times 10^{-11})/(9.0 \times 10^9)} = \sqrt{7.411 \times 10^{-21}} = 8.61 \times 10^{-11}$ .
With $m = 1.0$ kg, $q = 8.61 \times 10^{-11}\ \text{C}$ , that is about $86$ picocoulombs.
Convert to a count of electrons. EK 10.1.A.1.ii says the elementary charge $e$ can be considered to be the smallest indivisible amount of charge, and the sheets print $e = 1.60 \times 10^{-19}$ C. So $N = \dfrac{8.61 \times 10^{-11}}{1.60 \times 10^{-19}} = 5.38 \times 10^{8}$ electrons.
Weigh them. $(5.38 \times 10^8)(9.11 \times 10^{-31}) = 4.90 \times 10^{-22}\ \text{kg}$ . Against the sphere's $1.0$ kg, that is a mass change of about $4.9 \times 10^{-22}$ , roughly one part in $2 \times 10^{21}$ .
Read the result. Removing half a billion electrons sounds like a violent operation and is, by mass, undetectable. Ordinary matter is neutral to better than that, which is why the gravitational force between two kilogram masses is measurable at all. This is EK 10.1.B.3 as an arithmetic statement: gravitational forces dominate at larger scales because systems at large scales tend to be electrically neutral.

The gravitational force is $6.67 \times 10^{-11}$ N. With $1.0\ \mu\text{C}$ on each sphere the electric force is $9.0 \times 10^{-3}$ N, larger by a factor of $1.35 \times 10^8$ . The two match at a charge of only $8.61 \times 10^{-11}$ C on each, which is $5.38 \times 10^8$ electrons with a combined mass of $4.90 \times 10^{-22}$ kg, about one part in $2 \times 10^{21}$ of the sphere.

Same two spheres, three sign configurations

Two small spheres, each of mass $0.15$ kg, have centers $0.30$ m apart. Find the gravitational and electric forces, with directions, when (a) both carry $+2.0$ nC, (b) one carries $+2.0$ nC and the other $-2.0$ nC, and (c) both are neutral. Then find the charge magnitude at which the two forces would balance in case (a). Take toward the other sphere as the positive direction for each sphere.

Sign convention, declared first. For each sphere, positive means toward the other sphere. So an attractive force is positive and a repulsive force is negative.
Gravitational force, identical in all three cases. $\lvert \vec{F}_g \rvert = \dfrac{(6.67 \times 10^{-11})(0.15)(0.15)}{(0.30)^2} = \dfrac{(6.67 \times 10^{-11})(0.0225)}{0.0900} = (6.67 \times 10^{-11})(0.250) = 1.67 \times 10^{-11}\ \text{N}$ , attractive, so $+1.67 \times 10^{-11}$ N. Nothing done to the charges will change this line.
Case (a), both $+2.0$ nC. $\lvert \vec{F}_E \rvert = \dfrac{(9.0 \times 10^9)(2.0 \times 10^{-9})^2}{0.0900} = \dfrac{(9.0 \times 10^9)(4.0 \times 10^{-18})}{0.0900} = \dfrac{3.6 \times 10^{-8}}{0.0900} = 4.0 \times 10^{-7}\ \text{N}$ . Like signs repel by EK 10.1.A.3.i, so this is $-4.0 \times 10^{-7}$ N.
Case (a) net. $+1.67 \times 10^{-11} - 4.0 \times 10^{-7} = -4.0 \times 10^{-7}\ \text{N}$ to three figures. The gravitational term is $2.4 \times 10^4$ times smaller and does not survive the rounding.
Case (b), opposite signs. The magnitude of the electric force is unchanged, because Coulomb's law as printed uses $\lvert q_1 q_2 \rvert$ . Only the direction flips: opposite signs attract by EK 10.1.A.3.ii, so it is $+4.0 \times 10^{-7}$ N. Net: $+4.0 \times 10^{-7}$ N, attractive.
Case (c), both neutral. $q_1 q_2 = 0$ , so $\lvert \vec{F}_E \rvert = 0$ exactly. The gravitational force is still $+1.67 \times 10^{-11}$ N. One of the two forces has been switched off without touching the objects' positions or masses.
The balance charge for case (a). Setting the magnitudes equal, $q = m\sqrt{G/k} = (0.15)(8.61 \times 10^{-11}) = 1.29 \times 10^{-11}\ \text{C}$ . Check: $\dfrac{(9.0 \times 10^9)(1.29 \times 10^{-11})^2}{0.0900} = \dfrac{(9.0 \times 10^9)(1.667 \times 10^{-22})}{0.0900} = 1.67 \times 10^{-11}\ \text{N}$ , matching the gravitational value.
Compare with what they were given. $\dfrac{2.0 \times 10^{-9}}{1.29 \times 10^{-11}} = 155$ . The spheres were carrying $155$ times more charge than it takes for the electric force to match gravity, which is why case (a) was so lopsided.
Read the three cases as a table of what changed. Gravitational force: unchanged, unchanged, unchanged. Electric force: repulsive, attractive, zero. Only one of the two forces has any of those options available to it.

In all three cases the gravitational force is $1.67 \times 10^{-11}$ N, attractive. With both spheres at $+2.0$ nC the electric force is $4.0 \times 10^{-7}$ N repulsive; with opposite signs it is $4.0 \times 10^{-7}$ N attractive; neutral, it is exactly zero. The forces would balance at a charge of $1.29 \times 10^{-11}$ C, which is $155$ times less charge than the spheres were given.

Frequently asked questions

What is the difference between the gravitational force and the electric force?

Both are inverse-square laws that act along the line joining two objects, but three things separate them. Gravity is always attractive while the electric force attracts or repels depending on the signs of the charges. The electric force is enormously stronger for any pair of objects carrying meaningful charge. And charge can be cancelled to zero, screened by a conductor, and altered by the surrounding medium, while mass can do none of those things. The AP Physics 2 CED states the first three of these at essential knowledge 10.1.B.1 through 10.1.B.3, under a learning objective that asks students to describe the electric and gravitational forces that result from interactions between charged objects with mass.

How much stronger is the electric force than gravity?

There is no single number, because the answer depends on which two objects you pick. The ratio equals the Coulomb constant divided by the gravitational constant, multiplied by the ratio of the charge product to the mass product. Using the AP sheet values, two protons give about 1.2 times 10 to the 36th, and two electrons give about 4.2 times 10 to the 42nd. The six-order-of-magnitude gap between those comes entirely from the masses, and equals the square of the proton to electron mass ratio. What is fixed is the structure: because both forces fall off as one over r squared, the separation cancels out of the ratio entirely, so the answer is the same at any distance.

Does the ratio of electric force to gravitational force depend on distance?

No. Both laws have exactly the same distance dependence, one over r squared, so when you divide one by the other the r squared cancels and the ratio becomes a pure number set by the two objects' charges and masses. This is worth knowing because it closes off a common line of wrong reasoning, that gravity might catch up at large separations. It cannot. Gravity dominates at large scales for a different reason, which the CED gives at essential knowledge 10.1.B.3: systems at large scales tend to be electrically neutral, so their charges cancel while their masses add.

Why does gravity dominate at astronomical scales if it is so much weaker?

Because mass has one sign and charge has two. Every gram of a planet contributes to its gravitational pull and nothing subtracts, so the gravitational effect grows with the amount of matter without limit. Charge comes in positive and negative and cancels almost perfectly in bulk matter, so the electric effect of a planet is nearly zero no matter how big it is. Two further mechanisms keep it that way: excess charge drains away through grounding, described at essential knowledge 10.2.A.3, and conductors screen their interiors, described at essential knowledge 10.3.B.1. Essential knowledge 10.1.B.3 states the conclusion directly: gravitational forces dominate at larger scales even though they are weaker than electrostatic forces, because systems at large scales tend to be electrically neutral.

Can the electric force be shielded but gravity cannot?

Yes. Essential knowledge 10.3.B.1 in AP Physics 2 says that while in electrostatic equilibrium the excess charge of a solid conductor is distributed on the surface of the conductor, and the electric field within the conductor is zero. So a metal enclosure keeps external electric fields out of its interior. There is no material that does the equivalent for gravity, and none is expected. The electric force also depends on the medium the charges sit in, through the electric permittivity described at essential knowledge 10.1.C.1 and 10.1.C.4, while the gravitational force between two masses does not care what is between them.

Is Coulomb's law on the AP Physics 1 equation sheet?

No. The AP Physics 1 booklet prints the gravitational force law and the universal gravitational constant, but it carries no electricity table and no Coulomb constant, because AP Physics 1 does not cover the electric force. Coulomb's law is printed on the AP Physics 2 and AP Physics C: Electricity and Magnetism sheets, in both cases in the form with one over four pi epsilon nought and the equivalent k form, and both of those constants boxes print k as 9.0 times 10 to the ninth newton metre squared per coulomb squared. The gravitational force law is printed in the mechanics table of all four booklets, including the C: Electricity and Magnetism one, which carries a full mechanics table.

Is the normal force really the electric force in disguise?

Essentially yes, and the AP Physics 2 CED says so at essential knowledge 10.1.A.4: electric forces are responsible for some of the macroscopic properties of objects in everyday experiences, but the large number of particle interactions that occur make it more convenient to treat everyday forces in terms of nonfundamental forces called contact forces, such as normal force, friction, and tension. So the electric force is doing most of the pushing in the mechanics units as well, under other names. The word nonfundamental is the CED's, and it marks contact forces as bookkeeping devices rather than as separate interactions.