Electric Potential, Voltage, and Capacitance (4C) - MCAT Chemical and Physical Foundations of Biological Systems

$V$ is constant throughout the conductor. In electrostatic equilibrium, the potential is uniform inside and on the surface of a conductor due to zero internal field.

Perpendicular motion has $W=0$ (along equipotential). Motion perpendicular to the field lines follows equipotential surfaces, resulting in no change in potential and zero work.

$E=-\frac{dV}{dx}$. The electric field is the negative rate of change of potential with respect to position.

$U=\frac{kq_1q_2}{r}$. Represents the work to assemble the charges from infinity, analogous to gravitational potential energy.

$1\ \text{F}=1\ \text{C/V}$. The farad measures capacitance as one coulomb of charge stored per volt of potential difference.

$U=\frac{1}{2}CV^2=9\ \mu\text{J}$. Energy calculation uses the formula derived from integrating work done during charging.

$Q=CV=60\ \mu\text{C}$. Charge stored is the product of capacitance and applied voltage.

$C_{\text{eq}}=2C$. Parallel connection doubles effective plate area for identical capacitors, doubling capacitance.

$C=\frac{\varepsilon_0A}{d}$. For parallel plates, capacitance increases with area and decreases with separation, proportional to permittivity.

$U=\frac{Q^2}{2C}$. Alternative form obtained by substituting $Q=CV$ into the energy expression.

$C=\kappa\frac{\varepsilon_0A}{d}$. Dielectric insertion reduces the effective field, increasing capacitance by the factor $\kappa$.

$C_{\text{eq}}=\frac{C}{2}$. For identical capacitors in series, equivalent is half due to doubled effective separation.

$C_{\text{eq}}=\sum_i C_i$. In parallel, charges add while sharing the same voltage, summing individual capacitances.

$\frac{1}{C_{\text{eq}}}=\sum_i \frac{1}{C_i}$. In series, total voltage divides across capacitors, leading to reciprocal sum for equivalent capacitance.

$V=\frac{kQ}{r}$. Derived from integrating the electric field from infinity to $r$, assuming zero potential at infinity.

$\Delta U=q\Delta V$. Change in potential energy equals charge times the potential difference experienced by the charge.

$\Delta V=-\frac{W_{\text{field}}}{q}$. Potential difference equals the negative work done by the field per unit charge when moving a charge between points.

$V=\frac{U}{q}$ (electric potential energy per unit charge). Electric potential at a point is the electric potential energy per unit charge for a test charge placed there.

$U=\frac{1}{2}CV^2$. Energy stored derives from the work to charge the capacitor, integrating $Q dV$ from 0 to $V$.

$1\ \text{V}=1\ \text{J/C}$. The volt is defined as the potential difference that imparts one joule of energy to one coulomb of charge.

$\frac{V(2r)}{V(r)}=\frac{1}{2}$. Potential inversely proportional to distance, so doubling $r$ halves the potential.

$\Delta V=-E\Delta x$. In a uniform field, potential decreases linearly in the direction of the field by the product of field strength and distance.

$V<0$. For $Q<0$, potential is negative relative to zero at infinity, indicating attractive interaction for positive test charges.

$V_{\text{net}}=\sum_i \frac{kQ_i}{r_i}$. Electric potential is a scalar quantity, allowing direct summation of individual contributions.