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Take the conducting plane as $y=0$, with the real point charges $Q$ at $(0,b)$ and $-2Q$ at $(a,b)$. The method of images replaces the plane by image charges $-Q$ at $(0,-b)$ and $+2Q$ at $(a,-b)$. This makes the plane an equipotential; the physical field above it is the sum of the real- and image-charge fields. No factor of $1/2$ enters the force on a real charge.

Set $k_e=1/(4\pi {ϵ}_0)$ and $d=\sqrt{a^2+4b^2}$, and use positive $x$ from $Q$ toward $-2Q$, positive $y$ upward. The horizontal forces combine the mutual attraction and the force from the other charge's image:For the vertical components, the own-image attractions act downward; each cross-image term acts upward. ThusThese are the electrostatic forces; the external forces holding the charges fixed are their negatives. The smaller positive charge's vertical force need not always point downward, because the other charge's induced surface charge also contributes. The construction assumes the usual ideal large-plane limit with no imposed background field.

Let $C$ denote the capacitance omitted after the word “capacitance” in the source sentence. For the ideal series RC charging circuit, Kirchhoff's voltage law gives $RI+Q/C=V$, with $I=dQ/dt$ and $Q(0)=0$. ThereforeThe electric current falls from $V/R$ to zero, while the electric charge rises from zero to $CV$. Their initial slopes are $-V/(R\tau )$ and $V/R$, respectively.

Reducing $RC$ shortens the approach to the relevant final values. Which heights change depends on the component changed: reducing $R$ at fixed $C$ increases $I(0)$ but leaves $Q(\infty )$ unchanged; reducing $C$ at fixed $R$ leaves $I(0)$ unchanged but reduces $Q(\infty )$. A shorter time constant alone does not specify both endpoint amplitudes.

For normal incidence, the first-order maximum of a diffraction grating obeys $s\sin \theta =\lambda$. Differentiating at ${\lambda }_0$ givesTherefore the positive first-order peaks are separated byThe larger wavelength is diffracted farther from the normal. Both first orders must exist, ${\lambda }_0+\delta \lambda <s$; extremely close to grazing incidence the finite difference of the two inverse sines should replace this local approximation.

Let $\theta =\omega t+{\theta }_0$ be the angle between the coil's area normal and the uniform magnetic flux density. The linked magnetic flux is ${\Phi }_N=NBA\cos \theta$. Faraday's law givesThe sign uses the winding direction associated with the chosen normal; reversing the terminal convention changes it. The magnitude is largest when $\cos \theta =0$: the coil's normal is perpendicular to the magnetic field, so the plane of the coil is parallel to it. At that orientation the instantaneous magnetic flux is zero but its rate of change has maximum magnitude.

Choose the loop axis as $+x$, with positive electric current circulating by the right-hand rule. The Biot-Savart law isFor a loop element and an axial observation point, $r=\sqrt{a^2+x^2}$. The transverse components cancel between opposite elements. The axial component of $d\mathbf{\ell }\times \mathbf{r}$ is $ad\ell$, soThusReversing the electric current reverses the field. At the centre $B={\mu }_{0}I/(2a)$, while the far field decreases as $|x{|}^{-3}$, as expected for a magnetic dipole.