Electric field (E) & potential (V) are related as

A charge +Q at A as shown produces electric field E and electric potential V at D. If we now put charges - 3 Q and +Q at B and C, respectively, then the electric field and potential at D will be .

Electric field (E) and current density (J) have relation

Charges are placed on the vertices of a square as shown. Let E r be the electric field and V the potential at the centre. If the charges on A and B are interchanged with those on D and C respectively, then :-

A conductor with a cavity is charged positively and its surface charge density is sigma . If E and V represent the electric field and potential, then inside the cavity

A spherical charged conductor has surface charge density sigma .The intensity of electric field and potential on its surface are E and V .Now radius of sphere is halved keeping the charge density as constant .The new electric field on the surface and potential at the centre of the sphere are

In a uniform electric field, the potential is 10V at the origin of coordinates , and 8 V at each of the points (1, 0, 0), (0, 1, 0) and (0, 0, 1). The potential at the point (1, 1,1 ) will be .

Draw a plot showing the variation of (i) electric field (E ) and (ii) electric potential (V) with distance r due to a point charge Q.

Intensity of an electric field (E) depends on distance r due to a dipole, is related as

We know that electric field (E ) at any point in space can be calculated using the relation vecE = - (deltaV)/(deltax)hati - (deltaV)/(deltay)hatj - (deltaV)/(deltaz)hatk if we know the variation of potential (V) at that point. Now let the electric potential in volt along the x-axis vary as V = 2x^2 , where x is in meter. Its variation is as shown in figure Draw the variation of electric field (E ) along the x-axis.

The electric potential in a region is represented as V=2x+3y-z obtain expression for electric field strength.