$\mathrm{P}_1=\mathrm{q}(4)$

$\overrightarrow{\mathrm{P}}_1=\mathrm{P}_1 \hat{\mathrm{j}}$

$\overrightarrow{\mathrm{r}}=100(\hat{\mathrm{i}}+\hat{\mathrm{j}}) \mathrm{mm}$

$\mathrm{v}_0=\frac{K \mathrm{P}_1 \cdot \overrightarrow{\mathrm{r}}}{\mathrm{r}^3}=\frac{\mathrm{K}\left(100 \mathrm{P}_1\right)}{(100 \sqrt{2})^3}$

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$\tan \theta=2$

$\overrightarrow{\mathrm{P}}_2=\mathrm{P}_2[-\cos \theta \hat{\mathrm{i}}+\sin \theta \hat{\mathrm{j}}]$

$\overrightarrow{\mathrm{r}}=100(\hat{\mathrm{i}}+\hat{\mathrm{j}}) \mathrm{mm}$

$\mathrm{P}_2=\mathrm{q} \ell$

$\mathrm{v}=\frac{\mathrm{K} \overrightarrow{\mathrm{P}}_2 \cdot \overrightarrow{\mathrm{r}}}{\mathrm{r}^3}$

$\mathrm{v}=\frac{\mathrm{K}\left(100 \mathrm{P}_2\right)-(-\cos \theta+\sin \theta)}{(100 \sqrt{2})^3}$

$\frac{\mathrm{v}_0 \mathrm{P}_2}{\mathrm{P}_1}=(-\cos \theta+\sin \theta)$

$\mathrm{v}=\mathrm{v}_0 \frac{\mathrm{q} \ell}{\mathrm{q}(4)}[-\cos \theta+\sin \theta]$

$=\frac{\mathrm{v}_0}{4}[-2+4]=\frac{\mathrm{v}_0}{2}$