Electric Force, Field, and Potential

Heat to electric energy via thermionic emission due to light absorption raising temperature.

Chemical to electrical energy via photosynthesis-like reactions driven by light exposure.

Mechanical to electric energy via piezoelectric effect induced by photon pressure.

Electromagnetic radiation to electric energy via the photoelectric effect.

V = I/R because the voltage equals current divided by resistance.

V = IR because voltage is equal to the product of current and resistance.

R = VI because resistance is the product of voltage and current.

I = VR because current equals dividing voltage by resistance.

When it travels at any angle as long as both fields have equal strength.

When it travels parallel to both fields if they are also parallel to each other.

When it travels perpendicular to the electric field and parallel to the magnetic field.

When it travels perpendicular to both fields if they are parallel.

Within the dielectric material only

In the chemical bonds of capacitor materials

In the electric field between the plates

At one specific plate, either positive or negative

It remains constant because electrical forces are conservative.

It decreases because work is done by the electrostatic force.

It becomes zero as the particles are separated infinitely far apart.

It increases because work is done against the electrostatic force.

Activating bi-directional semiconductor junctions promote extensive carrier mobilizations which enhance local electrifications concomitantly preserving overall energetic equilibriums.

Transference efficiency peaks when reactive impedance summations approach nullification thresholds permitting uninhibited direct current flow throughout.

Harmonic oscillatory inductions maximize cross-sectional dielectric polarizations henceforth optimizing static retention capacities despite systemic resistances.

Incremental resistance additions cause insignificant dissipative heating effects thereby elevating static charging efficiencies notwithstanding alternative pathway options.

The voltage remains constant as it depends solely on battery voltage which hasn't changed.

The voltage across it increases due to $Q=CV$ relation where $Q$ is constant and $C$ decreases.

No definitive conclusion can be drawn without knowing how the dielectric material was altered to reduce capacitance.

The voltage decreases because reduced capacitance implies lower storage capability for charge at any given voltage level.

Ohmmeter

Wattmeter

Ammeter

Voltmeter

$W=Fd$

$P=IV$

$\epsilon=W/Q$

$P= I^2R$

$V = IR$

$E = qV$

$\Delta U + \Delta K = 0$

$P = IV$