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
A Gaussian surface in the figure is shown by dotted line. The electric field on
A Gaussian surface in the figure is shown by dotted line. The electric field on the surface will be:
- A. due to \( q_{1} \) and \( q_{2} \) only
- B. due to \( q_{2} \) only
- C. zero
- D. due to all
Step-by-step solution
According to the principle of superposition, the electric field at any point on the Gaussian surface is the vector sum of the fields due to all charges present (both inside and outside the surface). Gauss's law only relates the net flux through the surface to the enclosed charge, but does not restrict the field contribution to only those charges. Therefore, the electric field on the surface is due to all charges.
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