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a) The charge density at P increases and the electric field at T remains zero

b) The charge density at P remains zero and the electric field at T remains zero

c) The charge density at P increases. The electric field at T increases in magnitude and always points to the right

d) The charge density at P increases and the electric field at T remains zero

e) the charge density at P remains constant but non zero. The electric field at T remains constant, but points to the left

Answer
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In the figure 1 above a shell comprises a uniform positive charge on its surface. When a positively charged rod is brought close to it, the charge density moves on the other side due to electrostatic repulsion. This above process is called electrostatic induction.

A similar thing will take place when positive charge is kept at a distance r from the left side of the sphere. The charge density will shift towards the right of the sphere. Hence we can conclude that the charge density at point P will increase.

Now let us analyze the electric field at point T. For that let us say we enclose a Gaussian surface enclosing the inside the sphere such that the point T passes through the Gaussian surface. Hence by gauss law i.e. $\text{E}\text{. }\!\!\Delta\!\!\text{ S=}\dfrac{\text{q}}{{{\in }_{\circ }}}$ where E is the electric field on the Gaussian surface, $\Delta S$ is the surface area of the Gaussian surface, q is the charge enclosed within the surface and ${{\in }_{\circ }}$is the permittivity of free space. There is no charge present within the sphere. Hence q enclosed is zero which implies the electric field on the Gaussian surface is zero. This implies that the electric field at point P is also zero since point P lies on the Gaussian surface.