A Gaussian surface in the fig. is shown by dotted line. The electric field on the surface will be :-

A Gausisan surface in the figure is shown by dotted line. The electric field on the surface will be

A Gausisan surface in the figure is shown by dotted line. The electric field on the surface will be

A Gaussian surface encloses a proton p. The electric field at any point on the surface is vec€ . The flux linked with the Gaussian surface is phi . STATEMENT-1 : When an electron is kept close to this system outside the Gaussian surface, the flux linked with the surface would change. and STATEMENT-2 : The presence of electron will alterthe electric field on the gaussian surface.

A Gaussian surface encloses a proton p. The electric field at any point on the surface is vec€ . The flux linked with the Gaussian surface is phi . STATEMENT-1 : When an electron is kept close to this system outside the Gaussian surface, the flux linked with the surface would change. and STATEMENT-2 : The presence of electron will alterthe electric field on the gaussian surface.

S_1 : When a positively charged particle is released in an electric field , in its subsequent motion,it may or may not move along the electric field line passing through the point it has been released. S_2 : In electrostatic, conductors are always equipotential surfaces. S_3 : The flux through a closed Gaussian surface is non-zero. The electric field at some point on the Gaussian surface may be zero.

S_1 : When a positively charged particle is released in an electric field , in its subsequent motion,it may or may not move along the electric field line passing through the point it has been released. S_2 : In electrostatic, conductors are always equipotential surfaces. S_3 : The flux through a closed Gaussian surface is non-zero. The electric field at some point on the Gaussian surface may be zero.

Consider the charge configuration and a spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface, the electric field will be due to.