Electric Fields

01

Point-charge force and field

A source charge creates an electric field. The force on a chosen test charge depends on both that field and the test charge's sign and magnitude.

Use Coulomb's law with a direction

For two stationary point charges Q and q in free space or air, separated by distance r, the force acts along the line joining them. Its magnitude is:

Like signs repel; unlike signs attract. The force on Q and the force on q are equal and opposite, but act on different charges. They do not cancel in a force diagram for either one charge.

Here ε₀ is the permittivity of free space. Use the supplied rounded calculation value K = 9.00 × 10⁹ N m²/C². Charge is in coulombs, C, and r is in metres. One nanocoulomb, nC, is 10^-9 C.

Separate field from the test-charge force

Electric field strength is force per unit positive test charge, with direction given by the force on that positive charge. For a point source, its magnitude is K|Q|/r², directed away from positive Q and towards negative Q.

If outward from the source is chosen positive, the signed radial component and the test-charge force are:

In vector form, F = qE. A positive test charge experiences force along E; a negative one experiences force opposite E. A negative component has a meaning only after the positive direction has been stated.

Estimate the scale before substituting

A few nanocoulombs at a distance of a few tenths of a metre, with the supplied K, gives field of order 10²-10³ N/C. Acting on a nanocoulomb-scale charge, this suggests force of order 10^-6 N. The force is small even though the field's numerical value is hundreds, because their units and quantities differ.

Changing q alone changes its force, while the field due to the fixed source stays the same in this model. The test-charge approximation assumes the test charge does not rearrange the source. Doubling r quarters both the source field and the force on an unchanged q.

Check the permittivity units

The supplied permittivity is ε₀ = 8.85 × 10^-12 F/m. F here is the farad, defined in capacitance. Rearranging Coulomb's law gives:

This follows from C = A s and N = kg m/s². K is the reciprocal 4πε₀ factor. Separately rounded K and ε₀ may produce slightly different final calculator digits.

02

Electric potential and energy

Electric potential is energy per charge at a point. Electric potential energy belongs to the interaction of charges, and its change depends on the sign of the charge being moved.

Choose the zero reference

Electric potential at a point is the work done per unit small positive test charge by an external force in bringing it from infinity to that point, without changing its kinetic energy. Set potential to zero at infinity for the isolated point-source examples here.

Potential is a scalar. It is positive for a positive point source and negative for a negative source with this reference. For a positive test charge, moving slowly towards a positive source requires positive external work against repulsion. Moving it towards a negative source permits negative external work while the field does positive work.

A few nanocoulombs at a few tenths of a metre, with K of order 10¹⁰ N m²/C², suggests potential of order 10² V. The sign is a separate decision from that rough magnitude.

Multiply by the actual signed charge

For two point charges Q and q, the pair's electric potential energy, with zero at infinite separation, is:

U_E is in joules, whereas V is in joules per coulomb. Unlike charges have negative interaction energy relative to infinite separation: separating them without increasing kinetic energy requires an energy input. A negative energy is meaningful with the stated reference.

Potential difference is the change in potential between two points. For unchanged q:

If the field alone changes a particle's kinetic energy in the fixed-source model, neglecting radiation and other transfers, then ΔE_k = -ΔU_E = -qΔV. State whether motion is externally controlled or driven by the field alone before assigning work or a speed.

03

Field from a potential gradient

Electric field follows the local rate of decrease of potential. A potential graph's height, a chord gradient and a tangent gradient describe different quantities.

Use the negative local gradient

Along a chosen coordinate x, the field component is E_x = -dV/dx. For a point source with outward radial coordinate r:

A small radial movement dr has field work qE_rdr = -q dV, giving this relation. The field points towards decreasing potential. A negative charge's force points oppositely to the field.

The units agree because 1 V/m = 1 J/(C m) = 1 N/C. Field is a potential gradient, not the potential itself.

For the fixed +4.00 nC source with K = 9.00 × 10⁹ N m²/C², V(r) = 36/r when r is in metres and V in volts. A negative source of the same magnitude gives -36/r. Both tend to zero at indefinitely large r.

Equipotentials and zero points

Potential is constant along an equipotential. The electrostatic field is perpendicular to it; moving along it gives ΔV = 0 and therefore zero field work on a charge. A point with zero field does not automatically have zero potential.

Consider fixed sources at x = -0.300 and +0.300 m, with the common zero-potential reference at infinity. At their midpoint, add field vectors but add signed scalar potentials.

• Two +4.00 nC sources: fields are +400 and -400 N/C, so E = 0. Potentials are +120 and +120 V, giving V = +240 V.
• +4.00 nC on the left, -4.00 nC on the right: both fields point right, giving E = +800 N/C. Potentials +120 and -120 V sum to zero.

These results are consistent with E being a local gradient. A function can have a zero gradient at nonzero height, or pass through zero with a nonzero gradient.

04

Uniform fields between plates

Between ideal broad parallel plates, potential changes uniformly across the perpendicular gap. That gap determines the field, while the charge sign determines the force direction.

In the ideal interior region, neglecting edge effects, field magnitude is:

d is the perpendicular separation between plates. The field points from higher to lower potential. Plate length along a particle's path is a different distance and does not replace d.

The force is set by q and E, not by the velocity direction. If this is the only significant force on a particle of constant mass, it produces constant signed acceleration. Motion parallel to E and entry perpendicular to E therefore require different component descriptions.

05

Charged-particle motion

Calculate the force first, then resolve the motion. In a uniform field, acceleration is constant, but a component perpendicular to the field keeps its velocity if no other force acts.

Connect force to components

For constant mass with electric force alone, a = qE/m as a vector relation. A positive charge initially moving along E speeds up. A negative charge initially moving along E slows and may reverse if it remains in the field. For entry perpendicular to E, the initial velocity component along E is zero; that component then changes under the electric force, while the velocity component perpendicular to E stays constant.

Use the classical particle model below with a fixed electrostatic field. Neglect gravity, radiation and edge/fringing effects. Outside the specified ideal field region, assume no other force.

Inside the field, eliminating t gives y = a_yx²/(2u_x²), a parabola. After exit, acceleration is zero and both velocity components keep their exit values. The electron follows a straight line along the exit tangent; earlier acceleration does not keep bending it after the force ends.

Check the motion with energy

Potential increases upward with gradient 4000 V/m. For the calculated exit displacement:

The negative charge gains kinetic energy while moving to higher potential. Since its horizontal velocity is unchanged, the same gain is (1/2)m_ev_y,exit². Potential, potential energy and charge are distinct: the electron's charge has not changed.

06

Capacitance and plate charge

A capacitor stores charge separation between conductors. Its capacitance uses the charge magnitude on either plate, even when the two-plate system has zero net charge.

Define charge and potential difference locally

A capacitor has separated conductors that acquire opposite charges. For the usual initially neutral pair, one plate has +Q and the other -Q. Let V be the magnitude of their potential difference.

Capacitance C is charge stored divided by potential difference. Q is the magnitude on either plate, not the pair's algebraic net charge and not the sum of both magnitudes. The SI unit is the farad, F:

The C in the unit ratio is the coulomb; the quantity C is capacitance. Since a coulomb is A s and a volt is kg m² s^-3 A^-1, their ratio gives these base units. F as a unit means farad, while F as a force symbol has unit newton.

Useful conversions are 1 microfarad = 10^-6 F, 1 nanofarad = 10^-9 F and 1 microcoulomb = 10^-6 C. A rough 100 microcoulomb at 10 V suggests C of order 10 microfarads.

A linear capacitor has constant capacitance

For a given linear capacitor within its operating range, Q is proportional to V. C is then constant. This is a device-model property, not a guarantee that arbitrary Q/V pairs from every device agree.

For a measured record, identify how charge and voltage were determined, retain their units and resolutions, and examine whether the relation is linear over the tested range. The plotted values here are supplied model values, not experimental measurements. Reversing the graph axes changes the gradient to 1/C, as used in the energy calculation.

07

Energy stored in a capacitor

Separating charge requires work as the capacitor's potential difference grows. Its stored energy is the area under a potential-difference-against-charge graph.

Add the work of successive charge transfers

For a capacitor with plate-charge magnitude Q and potential difference V, transferring a small additional charge ΔQ requires work approximately VΔQ at that stage. Adding these small strips gives the energy stored in the ideal capacitor, with zero energy for the uncharged state.

The required graph has V vertically and Q horizontally. For a linear capacitor of constant C, V = Q/C is a straight line through the origin with gradient 1/C. The total area is a triangle:

These are three equivalent expressions for the same stored energy under the constant-capacitance model. Choose the form suited to the known or fixed quantities.

For charge of order 100 microcoulomb and voltage of order 10 V, QV/2 is of order 10^-3 J. One millijoule, mJ, is 10^-3 J.

State what stays fixed

For the same capacitor, doubling V doubles Q and quadruples U. The 10.0 microfarad device at 24.0 V therefore has Q = 240 microcoulombs and U = 2.88 mJ.

Comparing different capacitances requires a different decision. At fixed voltage U is proportional to C, whereas at fixed charge U is inversely proportional to C.

A disconnected isolated capacitor retains Q only under the ideal no-leakage assumption. A connected ideal source can maintain V by exchanging charge. The fixed-Q and fixed-V rows are separate comparisons; they do not hold both quantities fixed while changing C.

Distinguish source transfer from stored energy

For an initially uncharged linear capacitor charged through resistance from a constant-voltage source to its final voltage, the source transfers charge Q at source voltage V_s. Its energy transfer is V_sQ.

For the 12.0 V, 120 microcoulomb final state, the source transfers 1.44 mJ, while the capacitor stores 0.720 mJ. In this stated resistive charging model, the remaining 0.720 mJ is transferred in the resistance. The capacitor's changing voltage explains why its stored-energy area is triangular even though the source voltage stays constant.

This account depends on the initially uncharged capacitor, constant source voltage and stated charging process. Do not identify every source energy transfer with the capacitor's final stored energy.

08

Revision summary

State the source, test charge, direction and reference. Then select a force, field, potential or energy equation appropriate to that quantity.

Point charges: signs and inverse powers

These point-charge equations use free space or air and centre separation r. Like signs repel and unlike signs attract. The source determines E and V; test-charge sign then determines its force and pair energy. Potential is scalar, while fields and forces add as vectors.

Potential is external work per unit small positive test charge brought from infinity without a kinetic-energy change. With the isolated-source zero at infinity, V follows source sign and U_E follows the product of the two signs.

The work and motion statements require the stated fixed-source, negligible-other-transfer conditions. Moving -2.00 nC from +120 to +60.0 V raises its pair energy by +1.20 × 10^-7 J.

Use a local gradient and a perpendicular gap

A tangent slope of -400 V/m gives a +400 N/C component. A chord across a finite interval need not give that local value. Field is perpendicular to equipotentials and points towards decreasing V; a negative charge's force is opposite E.

At the midpoint between two equal positive sources, fields can cancel while potentials add. With equal opposite sources, potentials can cancel while fields reinforce. Zero E and zero V are different conditions.

For plates at +120 and 0 V separated by 0.0300 m, the field is 4000 V/m from the higher to lower potential. Use the perpendicular gap, not plate length. A chosen 0 V label does not imply an Earth connection.

Calculate motion in components

With electric force alone, a = qE/m. An electron is accelerated opposite E. In the supplied transverse-entry model, x = u_xt and y = (1/2)a_yt², with v_x unchanged. Transit is 3.00 ns, exit deflection is 3.16 mm and exit v_y is 2.11 × 10⁶ m/s upward. This is inside the 15.0 mm half-gap.

After the ideal field ends, force-free motion follows the exit tangent. It does not continue curving or reset to horizontal. Revisit the component graphs and energy check to connect the exit state to its assumptions.

Capacitor charge, graph axes and constraints

Q is the magnitude on either plate. Opposite charges +Q and -Q give zero net pair charge while storing charge separation. Constant C describes a linear capacitor within its operating range.

Q against V has gradient C. V against Q has gradient 1/C, and its area gives stored energy. For the 10.0 microfarad, 12.0 V model, Q = 120 microcoulombs and U = 0.720 mJ.

• Same C, doubled V: Q doubles and U quadruples.
• Same V, doubled C: Q and U double.
• Same Q, doubled C: V and U halve.

State whether a source maintains V or an isolated, non-leaking capacitor retains Q. In the specified initially uncharged, constant-voltage resistive charging model, source transfer V_sQ exceeds the final stored energy; account for energy transferred in the resistance.

Quantities and units

Use 10^-9 for nano, 10^-6 for micro and 10^-3 for milli. Here E means field; energy is U or E_k. The letter C can mean capacitance as a quantity or coulomb as a unit, while F can mean force as a quantity or farad as a unit.