. . . potential.¹

In case this is not self-evident, remember that the force $\vec{F}$ on a charge q is given by $\vec{F}$= q$\vec{E}$, where $\vec{E}$ is the electric field, measured in N/C or V/m. If we move a particle a distance d$\vec{x}$ under the influence of a force $\vec{F}$, that force does $\vec{F} \cdot$d$\vec{x}$  work on the particle - which appears as kinetic energy! If we move a charge q=e a distance x parallel to a constant electric field E, the work done is e E x, or (e)(E [V/m])(x [m]) = eV. After all, the reason $\phi$ is called the ``electric potential'' is that when multiplied by q it gives the potential energy of the charge in the electric field.

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. . . kilowatt-hour?²

Recall that watts (W) are a unit of power (energy per unit time) equal to joules per second: 1 W $\equiv$ 1 J/s.

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. . . drawings,³

For this and the next question, the final result is well known and the full derivation can be found in any textbook; however, you will not really own GAUSS' LAW until you have expressed and used it your own way. This is true for almost everything, but GAUSS' LAW is so important (and so simple!) that you should not miss this chance to make it yours.

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. . . WOOPS!⁴

We forgot to hand out last week's assignment on Thermodynamics! These were the problems; see if you can find time to do them over then next few weeks - no deadline, but it would be wise to have a look before the Physics Midterm.

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