Asked by xochitl

Explain how the charges on objects determine the electric force between them.

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Answered by GPT-5 mini AI
The electric force between charged objects is set by three main things: how much charge they carry (magnitude and sign), how far apart they are, and the medium between them. Key points:

- Like charges repel, opposite charges attract. The sign of the charges determines the direction (attraction or repulsion); the magnitudes determine how strong the force is.

- Coulomb’s law (for point charges) gives the strength and direction:
F = k |q1 q2| / r^2 (magnitude),
and in vector form F12 = k (q1 q2 / r^2) r̂12,
where q1 and q2 are the charges, r is the separation, r̂12 is the unit vector from charge 1 toward charge 2, and k = 1/(4πϵ0) ≈ 8.99×10^9 N·m^2/C^2. The product q1 q2 carries the sign: positive product → repulsive, negative → attractive.

- Distance matters strongly: force falls off as 1/r^2. Double a separation and the force becomes one-quarter.

- Linear dependence on each charge: doubling either q1 or q2 doubles the force.

- Superposition principle: the net force on a charge equals the vector sum of the forces from each other charge acting independently.

- The medium changes the force: in a dielectric the force is reduced by the material’s relative permittivity εr (replace k with 1/(4πϵ0εr)).

- For extended objects, you must consider the distribution of charge. If they are small compared with their separation, you can treat them as point charges; otherwise integrate the contributions of all charge elements.

Example: q1 = 1.0×10^-6 C, q2 = 2.0×10^-6 C, r = 0.10 m → F ≈ (8.99×10^9)(2×10^-12)/(0.01) ≈ 1.8 N (repulsive if both charges are positive).

In summary: the signs tell you whether the force is attractive or repulsive, the magnitudes and distance determine how large the force is (via Coulomb’s 1/r^2 law), and you add contributions vectorially when many charges are present.