JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2019
- Hard
- MCQ
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 \({\mkern 1mu} {\mkern 1mu} \frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}{\mkern 1mu} {\mkern 1mu} N{m^2}{C^{ - 2}}{\rm{ }}\) ]
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \((63\hat i - 27\hat j) \times {10^2}\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow {\text{E}} = \frac{{{\text{k}}{{\text{q}}_1}}}{{{\text{r}}_1^3}}{\overrightarrow {\text{r}} _1} + \frac{{{\text{k}}{q_2}}}{{{\text{r}}_2^3}}{\overrightarrow {\text{r}} _2}\)
\( = {\text{k}} \times {10^{ - 6}}\left[ {\frac{{\sqrt {10} }}{{10\sqrt {10} }}( - \widehat {\text{i}} + 3\widehat {\text{j}}) + \frac{{( - 25)}}{{125}}( - 4\widehat {\text{i}} + 3\widehat {\text{j}})} \right]\)
\( = \left( {9 \times {{10}^3}} \right)\left[ {\frac{1}{{10}}( - \widehat {\text{i}} + 3\widehat {\text{j}}) - \frac{1}{5}( - 4\widehat {\text{i}} + 3\widehat {\text{j}})} \right]\)
\( = \left( {9 \times {{10}^3}} \right)\left[ {\left( { - \frac{1}{{10}} + \frac{4}{5}} \right)\widehat {\text{i}} + \left( {\frac{3}{{10}} - \frac{3}{5}} \right)\widehat {\text{i}}} \right]\)
\( = 9000\left( {\frac{7}{{10}}\widehat {\text{i}} - \frac{3}{{10}}\widehat {\text{j}}} \right)\)
\( = (63\widehat {\text{i}} - 27\widehat {\text{j}})(100)\)
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