# Question d3ad2

May 22, 2017

B) Electric field remains unchanged

#### Explanation:

The change in the magnitude of the test charge would not affect the strength of the electric field which is determined by the magnitude of the source charge $\left({Q}_{s}\right)$ and the distance from it $\left({r}^{2}\right)$.

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There are essentially two ways that you can figure out the electric field.

color(white)(----)color(magenta)(vecE = (kQ_(s))/r^2)color(white)(--)orcolor(white)(--)color(orange)(vecE = F/q_(t)

Where
$\text{k = Coulomb's constant} \left(9 \times {10}^{9} \frac{N \cdot {m}^{2}}{{C}^{2}}\right)$
${Q}_{s} = \text{source charge (C)}$
$\text{r = distance from the two charges (m)}$
$\text{F = force (N)}$
${q}_{t} = \text{test charge (C)}$

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$\textcolor{red}{\text{A) Wrong}}$ The electric field would not double because the electric field is set up by the source charge, and therefore, any changes made to the magnitude of the test charge would have no effect.

$\textcolor{red}{\text{B) Wrong}}$ The electric field, again, would not change due to a change in the magnitude of the test charge. If the magnitude of the test charge did change, since the electric field remains constant, the Force would have to increase by the same magnitude.

Mathematically, this is seen as follows:

Example: If ${q}_{t}$ doubles then...?**

color(white)(--)stackrel"(constant)"vecE = [Fxx(2)]/[q_(t)xx(2)] ..."the Force would also have to double to keep the ratio constant"#