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Physics · JEE | Electric Charges and Fields

हिन्दी

What is the expression for the electric field at position vector \( \mathbf{r} \) due to a system of discrete charges \( q_1, q_2, \dots, q_n \) located at position vectors \( \mathbf{r}_1, \mathbf{r}_2, \dots, \mathbf{r}_n \)?

  1. A
    \( \mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \sum_{i=1}^{n} \frac{q_i}{r_{iP}} \hat{\mathbf{r}}_{iP} \)
  2. B
    \( \mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \sum_{i=1}^{n} \frac{q_i}{r_{iP}^2} \hat{\mathbf{r}}_{iP} \)
  3. C
    \( \mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \sum_{i=1}^{n} \frac{q_i^2}{r_{iP}^2} \hat{\mathbf{r}}_{iP} \)
  4. D
    \( \mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \sum_{i=1}^{n} \frac{q_i}{r_{iP}^3} \hat{\mathbf{r}}_{iP} \)
View Answer & Solution

Correct answer · Option B

\( \mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \sum_{i=1}^{n} \frac{q_i}{r_{iP}^2} \hat{\mathbf{r}}_{iP} \)

Explanation

From the superposition principle, the total electric field is the vector sum of individual electric fields: \( \mathbf{E} = \frac{1}{4\pi\varepsilon_0}\sum \frac{q_i}{r_{iP}^2}\hat{\mathbf{r}}_{iP} \).
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