Online Electricity & Magnetism Tutoring

Field and force are not interchangeable

Let $\mathbf E$ denote a uniform electric field pointing right. A small test charge has signed charge $q<0$. In which direction does the electric force $\mathbf F$ on that charge point?

Work through the reasoning

The relation is $\mathbf F=q\mathbf E$: multiply the field vector by the signed charge. Multiplication by a negative number reverses a vector’s direction, so the force points left. Its magnitude is $|q|\,\lVert\mathbf E\rVert$, where $|q|$ is the absolute value of the charge and $\lVert\mathbf E\rVert$ is the field vector’s magnitude. The field describes the influence of the source charges at the observation point; the force also depends on the test charge placed there. A positive test charge would feel a force to the right in this same field. Draw those two cases before doing longer component calculations.

Prepare for your lesson

Bring a diagram that separates the source charges, the observation point and the test charge. Label the field direction before finding the force direction. Keep the charge’s sign when multiplying a field vector, and use the absolute value only when calculating a magnitude.

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The connection chooses what stays fixed

An ideal 2 µF capacitor with no leakage is charged to 6 V. Its capacitance is then changed to 4 µF. Compare two separate cases: it stays connected to an ideal 6 V source, or it is disconnected first and remains isolated. Which quantity stays fixed, and what are the final charge and stored energy?

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