Q: Step 3: Square the distance.

The distance $r$ is $0.04 \text{ m}$. Squaring this value ($0.04 \times 0.04$) gives $r^2 = 0.0016 \text{ m}^2$, which can also be written in scientific notation as $1.6 \times 10^{-3} \text{ m}^2$.

Q: Step 5: Multiply by Coulomb's constant.

Finally, multiply this quotient by $k$. $F = (8.99 \times 10^9) \cdot (9.375 \times 10^{-9})$. Notice how the $10^9$ and $10^{-9}$ exponents effectively cancel each other out ($10^0 = 1$). $F = 8.99 \cdot 9.375 = 84.28125 \text{ Newtons}$.

Q: Step 6: Determine the direction based on physical principles.

The magnitude of the force is approximately $84.3 \text{ N}$. Because Sphere A is positive and Sphere B is negative, they possess opposite charges. According to the fundamental rules of electrostatics, opposites attract. Therefore, the final answer is an attractive force of $84.3 \text{ Newtons}$ pulling the spheres toward each other.

Question 1

Step 1: Convert all units to standard SI units.

Accepted Answer

This is where most beginners fail.

Question 2

Step 2: Apply the absolute value to the charges.

Accepted Answer

We only care about the magnitude of the force right now. Therefore, $|q_1 \cdot q_2| = (3.0 	imes 10^{-6}) \cdot (5.0 	imes 10^{-6})$. Multiplying these together yields $15.0 	imes 10^{-12} 	ext{ C}^2$.

Question 3

Step 3: Square the distance.

Accepted Answer

The distance $r$ is $0.04 	ext{ m}$. Squaring this value ($0.04 	imes 0.04$) gives $r^2 = 0.0016 	ext{ m}^2$, which can also be written in scientific notation as $1.6 	imes 10^{-3} 	ext{ m}^2$.

Question 4

Step 4: Divide the charge product by the squared distance.

Accepted Answer

Take the result from Step 2 and divide it by the result from Step 3. 
$(15.0 	imes 10^{-12}) / (1.6 	imes 10^{-3}) = 9.375 	imes 10^{-9} 	ext{ C}^2/	ext{m}^2$.

Question 5

Step 5: Multiply by Coulomb's constant.

Accepted Answer

Finally, multiply this quotient by $k$.
$F = (8.99 	imes 10^9) \cdot (9.375 	imes 10^{-9})$.
Notice how the $10^9$ and $10^{-9}$ exponents effectively cancel each other out ($10^0 = 1$). 
$F = 8.99 \cdot 9.375 = 84.28125 	ext{ Newtons}$.

Question 6

Step 6: Determine the direction based on physical principles.

Accepted Answer

The magnitude of the force is approximately $84.3 	ext{ N}$. Because Sphere A is positive and Sphere B is negative, they possess opposite charges. According to the fundamental rules of electrostatics, opposites attract. Therefore, the final answer is an attractive force of $84.3 	ext{ Newtons}$ pulling the spheres toward each other.

Question 7

Can the electrostatic force calculated by Coulomb's Law be negative?

Accepted Answer

Yes, but it depends entirely on how you are applying the formula. If you are using the basic scalar formula ($F = k |q_1 q_2| / r^2$) with absolute values, the force will always be a positive number representing pure magnitude. However, if you remove the absolute value bars, multiplying a positive charge by a negative charge yields a negative force. In physics convention, a negative force result indicates an attractive force (pulling together), while a positive force result indicates a repulsive force (pushing apart).

Question 8

What happens to the force if the distance between the charges is cut in half?

Accepted Answer

Because Coulomb's Law is an inverse-square law, the force increases exponentially as the distance decreases. If you cut the distance ($r$) in half, the denominator of the equation becomes $(1/2)^2$, which is $1/4$. Dividing by $1/4$ is mathematically identical to multiplying by $4$. Therefore, halving the distance makes the electrostatic force exactly four times stronger.

Question 9

Why do we use microcoulombs ($\mu	ext{C}$) instead of regular Coulombs in most problems?

Accepted Answer

A single Coulomb is an extraordinarily massive quantity of concentrated electrical charge, representing roughly $6.24 	imes 10^{18}$ individual electrons. If you held a sphere with 1 Coulomb of positive charge and placed it 1 meter away from a sphere with 1 Coulomb of negative charge, the attractive force between them would be $8.99 	imes 10^9$ Newtons—roughly equal to the weight of a million metric tons. Real-world static electricity scenarios involve vastly smaller amounts of charge, making the microcoulomb ($10^{-6} 	ext{ C}$) a much more practical unit of measurement.

Question 10

Does Coulomb's Law apply in water or other liquids?

Accepted Answer

Yes, but the force is significantly reduced. Coulomb's standard constant ($k = 8.99 	imes 10^9$) assumes a vacuum or dry air. Water is a dielectric material; its molecules are polar and align themselves to oppose the electric field between the charges. To calculate the force in water, you must divide the standard vacuum force by water's dielectric constant, which is approximately 80. This means the electrostatic force between two charges in water is 80 times weaker than it would be in an empty vacuum.

Question 11

How do you calculate the force when there are three or more charges?

Accepted Answer

You cannot calculate the force of three charges simultaneously in a single equation. You must use the Principle of Superposition. Pick the specific charge you want to analyze. Calculate the force exerted on it by the second charge using Coulomb's Law. Then, calculate the force exerted on it by the third charge using Coulomb's Law again. Finally, use vector addition (breaking the forces into X and Y trigonometric components) to add the individual force vectors together to find the total net force and direction.

Question 12

What is the difference between Coulomb's constant ($k$) and the permittivity of free space ($\epsilon_0$)?

Accepted Answer

They are two different mathematical ways of expressing the exact same physical property of the universe. The permittivity of free space ($\epsilon_0 \approx 8.854 	imes 10^{-12}$) describes how easily an electric field can penetrate a vacuum. Coulomb's constant ($k$) is simply a mathematical combination defined as $k = 1 / (4\pi\epsilon_0)$. Physicists use $k$ in Coulomb's Law because it makes the algebraic equation much cleaner and easier to write. They use $\epsilon_0$ in more advanced formulas, like Gauss's Law, where the $4\pi$ geometry naturally cancels out.

Question 13

Is the electrostatic force stronger than gravity?

Accepted Answer

Yes, by an almost incomprehensible margin. If you compare the electrostatic repulsion between two protons to the gravitational attraction between their masses, the electrostatic force is roughly $10^{36}$ times stronger. The only reason gravity dominates the macroscopic universe (holding planets and stars together) is that most large objects are electrically neutral, meaning their positive and negative charges cancel out perfectly. Gravity has no negative counterpart to cancel it out, so its weak force accumulates over massive scales.

Coulomb's Law Calculator

Command Palette