$A$ point charge of magnitude $+ 1\,\mu C$ is fixed at $(0, 0, 0)$. An isolated uncharged spherical conductor is fixed with its center at $(4, 0, 0)$. The potential and the induced electric field at the center of the sphere are:

'At the surface of a charged conductor,the electrostatic field must be normal to the surface at every point'. Explain.

The charge given to any conductor resides on its outer surface,because

Three identical uncharged metal spheres are at the vertices of an equilateral triangle. One at a time,a small sphere is connected by a conducting wire with a large metal sphere that is charged. The center of the large sphere is on the straight line perpendicular to the plane of the equilateral triangle and passing through its center (see figure). As a result,the first small sphere acquires charge $q_1$ and the second acquires charge $q_2$ $(q_2 < q_1)$. The charge that the third sphere $q_3$ will acquire is (Assume $l >> R$,$l >> r$,$d >> R$,$d >> r$):

Conduction electrons are almost uniformly distributed within a conducting plate. When placed in an electrostatic field $\overrightarrow{E}$,the electric field within the plate

Two spheres $A$ and $B$ are kept very near to each other. $A$ is negatively charged and $B$ is earthed. The true statement is:
$(A)$ Charge on $B$ is zero
$(B)$ Potential at $B$ is zero
$(C)$ Charge is uniformly distributed on $A$
$(D)$ Charge is non-uniformly distributed on $A$