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Put $+q$ at $x=0$ and $+4q$ at $x=l$, and write $k_e=1/(4\pi {ϵ}_0)$. The electric fields of the two positive point charges can cancel only between them. For $Q\ne 0$, zero Lorentz force on $Q$ therefore requiresThe force of $+4q$ on $+q$ points left, so $Q$ must be negative to pull $+q$ right. Equating these forces givesFor $+4q$, the attraction to $Q$ at separation $2l/3$ has magnitude $k_e(4q)(4q/9)/(2l/3{)}^2=4k_e{q}^2/l^2$, exactly balancing its repulsion from $+q$. Thus all three forces vanish, not only the force on the inserted charge.

The ideal voltage source drives a series $L$–$R$–$C$ branch and a separate resistor $R$ in parallel. At ${\omega }_0=1/\sqrt{LC}$, the series branch is purely resistive, so its peak current is $I_p=V_0/R$. Its stored energies areTheir sum is constant in sinusoidal steady state, with value $U_{\max }=L{V}_0^2/(2R^2)$. Each resistor dissipates average power $V_0^2/(2R)$, including the resistor directly across the source. Over one cycle,Omitting the parallel resistor would double this ratio. The expression concerns the driven steady state, not a switch-on transient.

Take B as reference and define positive source current $I$ upward into the upper internal node; the source symbol does not specify an arrow, so this definition fixes its sign. With A open, no current flows through the upper series resistor. The shunt resistor therefore has voltage $IR$, and the battery's longer plate is on the right, adding $+V$ toward A. ThusDeactivate the independent sources: open the ideal current source and short the ideal voltage source. The two resistors then lie in series between A and B, givingThe Thévenin equivalent is a source $IR+V$ in series with $2R$, positive toward A. If $I$ is instead defined downward, replace $I$ by $-I$; the resistance is unchanged.

In classical Maxwell's equations, electric charge is a source or sink of the electric field, so in vacuum $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$ can be nonzero. There is no magnetic-charge source in the same theory: Gauss's law for magnetism is $\nabla \cdot \mathbf{B}=0$. Magnetic flux entering a closed surface equals the flux leaving it. A magnetic dipole has both polarities; it is not an isolated magnetic monopole.