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Two point charges $q_1\,(\sqrt {10}\,\,\mu C)$ and $q_2\,(-25\,\,\mu C)$ are placed on the $x-$ axis at $x = 1\,m$ and $x = 4\,m$ respectively. The electric field (in $V/m$ ) at a point $y = 3\,m$ on $y-$ axis is, [ take ${\mkern 1mu} {\mkern 1mu} \frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}{\mkern 1mu} {\mkern 1mu} N{m^2}{C^{ - 2}}{\rm{ }}$ ]
$(63\hat i - 27\hat j) \times {10^2}$
$(-63\hat i + 27\hat j) \times {10^2}$
$(81\hat i - 81\hat j) \times {10^2}$
$(-81\hat i + 81\hat j) \times {10^2}$
Solution
$\overrightarrow {\text{E}} = \frac{{{\text{k}}{{\text{q}}_1}}}{{{\text{r}}_1^3}}{\overrightarrow {\text{r}} _1} + \frac{{{\text{k}}{q_2}}}{{{\text{r}}_2^3}}{\overrightarrow {\text{r}} _2}$
$ = {\text{k}} \times {10^{ – 6}}\left[ {\frac{{\sqrt {10} }}{{10\sqrt {10} }}( – \widehat {\text{i}} + 3\widehat {\text{j}}) + \frac{{( – 25)}}{{125}}( – 4\widehat {\text{i}} + 3\widehat {\text{j}})} \right]$
$ = \left( {9 \times {{10}^3}} \right)\left[ {\frac{1}{{10}}( – \widehat {\text{i}} + 3\widehat {\text{j}}) – \frac{1}{5}( – 4\widehat {\text{i}} + 3\widehat {\text{j}})} \right]$
$ = \left( {9 \times {{10}^3}} \right)\left[ {\left( { – \frac{1}{{10}} + \frac{4}{5}} \right)\widehat {\text{i}} + \left( {\frac{3}{{10}} – \frac{3}{5}} \right)\widehat {\text{i}}} \right]$
$ = 9000\left( {\frac{7}{{10}}\widehat {\text{i}} – \frac{3}{{10}}\widehat {\text{j}}} \right)$
$ = (63\widehat {\text{i}} – 27\widehat {\text{j}})(100)$
