Charge interaction and electric field

Two charged spheres deflect more when brought closer. How can we predict the force magnitude, then describe the action of one charge even where the second sphere has not yet been placed?

First establish direction. For two stationary point charges, forces act along the line joining them: like signs repel, unlike signs attract. Each object experiences its own force; their magnitudes are equal and directions opposite. They must not be added as two forces on one sphere. Electric force F is measured in newtons. Signs determine direction, but force magnitude also requires distance and the amounts of charge.

Can a charged sphere be replaced by one point? Yes, if its size is small compared with distance r to the other object and details of charge distribution are unimportant. This is the point-charge model; r is measured in metres between the chosen points. It separates a simple relationship from object shape. In nearby, complex conductors, charges redistribute, so putting centre-to-centre distance into the simple formula may be incorrect.

Keep charges unchanged and double their separation. The experimental relation for point charges gives one quarter of the force; tripling the distance gives one ninth. This is an inverse-square relation, not a relation inversely proportional to r itself. Greater separation weakens the interaction. Force grows on approach, but real spheres cannot be taken to zero separation within this model: their finite sizes become important first.

Combine the dependences: doubling one charge doubles force, doubling both quadruples it, and distance appears squared in the denominator. Coulomb's law expresses this as F = k|q₁q₂|/r². In vacuum, k ≈ 9·10⁹ Н·м²/Кл². Charges in C and r in m give force in N. If either charge is zero, this point-charge interaction vanishes. The absolute product gives force magnitude; determine direction separately.

Interaction also depends on the medium: its charged particles respond to the field. The vacuum coefficient therefore cannot be used unchanged for every liquid or solid; air can usually be approximated as vacuum. Convert charges before calculating: 1 мкКл = 10⁻⁶ Кл, 1 нКл = 10⁻⁹ Кл. Squaring distance also squares its units: convert centimetres to metres first, then square the number.

Rather than recalculate the interaction for each sphere, describe what the first charge creates at every surrounding point: an electric field. Another charge placed at a point experiences electric force. Imagine moving a small positive test charge to investigate. Field direction is defined as the force direction on that charge. The field map thus describes the source's action before a particular test object is chosen.

At the same field location, a larger test charge experiences more force. To separate the field's property from the test charge's property, divide force by charge magnitude: E = F/|qпроб|. This is electric field magnitude, in N/C. Double a small test charge: F doubles but the ratio stays unchanged. Reduce it: force falls but E does not vanish. The test charge must be nonzero and small enough to barely disturb the charge distribution being studied.

What would a negative test charge show? For the same charge magnitude, force reverses, but the source field does not. Positive charge feels force along the field; negative charge feels it against the field. Electrostatic field lines leave positive charges, enter negative charges or extend to infinity. They show force direction on a positive test charge, not compulsory paths for every charge's motion.

Controlling charges and fields can direct small particles towards the collecting surfaces of an electrostatic precipitator. Accumulated charge can also create a strong field and a spark discharge. Near flammable vapours, a spark may ignite them. Purpose-designed grounding gives charge a path to drain away, reducing accumulation. The same field physics explains both the useful device and its design requirements.

A field exerts a force of 0.006 N to the right on a +2 μC test charge. Find the electric field strength.

Find the field property by dividing electric force by test-charge magnitude. First convert +2 μC to +0.000002 C.

E = 0,006/0,000002 = 3000 Н/Кл. The force-to-charge-magnitude ratio for a small test charge at this point is 3000 N per coulomb.

The test charge is positive, so the field points right, in the stated force direction. Check: 3000·0,000002 = 0,006 Н.

E = 3000 Н/Кл to the right. A negative test charge of the same magnitude would experience 0.006 N to the left.

Coulomb's law links charge magnitudes, separation and force for stationary point charges.

Field strength characterizes the field at a point; dividing by test charge removes dependence on the chosen probe.

Field direction is defined by a positive test charge; force on a negative charge is opposite.

When a field acts on mobile charges in a conductor, electric current can flow.