The potential at a point x ( m) due to some charges situated on the X - axis is given by V(x) = x^2 + 1.5 x + 2 volt The electric field E at x = 3 m is given by

Step-by-step solution

The electric potential is given by V(x) = x^2 + 1.5 x + 2 The electric field is related to the potential by the equation E = - dV dx Differentiating the given potential with respect to x : E = - d dx (x^2 + 1.5 x + 2) = -(2x + 1.5) Substituting x = 3 m : E = -(2(3) + 1.5) = -7.5 V / m The magnitude of the electric field is 7.5 V / m and the negative sign indicates that it is directed in the negative X-direction. Answer: 7.5 V / m and in the -ve X - direction.