Physics · Electric flux, Gauss's law and its applications to find field due to infinitely long uniformly charged straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell
If the electric flux entering and leaving a closed surface are and units respect
If the electric flux entering and leaving a closed surface are \( 6 \times 10^{6} \) and \( 9 \times \) \( 10^{6} S I \) units respectively, then the charge inside the surface of permittivity of free space \( \omega_{0} \) is
- A. \( \omega_{0} \times 10^{6} \)
- B. \( -\omega_{0} \times 10^{6} \)
- C. \( -2 \omega_{0} \times 10^{6} \)
- D. \( 3 \omega_{0} \times 10^{6} \) E \( .2 \omega_{0} \times 10^{6} \)
Step-by-step solution
Using Gauss's law, net electric flux through a closed surface equals charge enclosed divided by permittivity of free space. Flux entering is negative, flux leaving is positive. Net flux = 9×10^6 - 6×10^6 = 3×10^6 SI units. Therefore, charge enclosed = ε0 × 3×10^6 = 3 ε0 × 10^6, which is option D.
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