JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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 \({\mkern 1mu} {\mkern 1mu} \frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}{\mkern 1mu} {\mkern 1mu} N{m^2}{C^{ - 2}}{\rm{ }}\) ]
- A \((63\hat i - 27\hat j) \times {10^2}\)
- B \((-63\hat i + 27\hat j) \times {10^2}\)
- C \((81\hat i - 81\hat j) \times {10^2}\)
- D \((-81\hat i + 81\hat j) \times {10^2}\)
Answer & Solution
Correct Answer
(A) \((63\hat i - 27\hat j) \times {10^2}\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow {\text{E}} = \frac{{{\text{k}}{{\text{q}}_1}}}{{{\text{r}}_1^3}}{\overrightarrow {\text{r}} _1} + \frac{{{\text{k}}{q_2}}}{{{\text{r}}_2^3}}{\overrightarrow {\text{r}} _2}\)…
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