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# A Gaussian surface encloses two of the four positively charged particles. The particles that contribute to the electric field at the point P on the surface are:A. ${{q}_{1}}\;and\;{{q}_{2}}$ B. ${{q}_{2}}\;and\;{{q}_{3}}$C. ${{q}_{1}}\;and\;{{q}_{3}}$D. ${{q}_{1}}\;,{{q}_{2}},{{q}_{3}}\;and\;{{q}_{4}}$

Last updated date: 12th Aug 2024
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Hint: Use the expression for electric field due to a discrete charge distribution.

Formula used: $\overrightarrow{E}=\sum\limits_{i=1}^{4}{\dfrac{{{q}_{i}}}{4\pi {{\epsilon }_{0}}r_{i}^{3}}\overrightarrow{{{r}_{i}}}}$
$\overrightarrow{E}$ is the electric field at the point P due to all the charges
${{q}_{i}}$ are the four charges
$\epsilon_0$ is the permittivity of free space
$\overrightarrow{{{r}_{i}}}$ is the position vector of the point P with respect to the ${{q}_{i}}$ charge

Complete step by step solution:
According to the formula, the electric field at any point is due to the cumulative effect of all the charges present irrespective of the Gaussian surface. Hence, all the four charges ${{q}_{1}},{{q}_{2}},{{q}_{3}}\;and\;{{q}_{4}}$ will contribute to the electric field.

The correct answer is option D.