If potential (in volt) in a region is expressed as V ( x , y , z )=6 xy - y +2 yz , the electric field (in N C⁻¹ ) at point (1,0,1) is

Step-by-step solution

The electric field E is related to the potential V by the relation E = - V = - ( V x i + V y j + V z k ) . Given V(x, y, z) = 6xy - y + 2yz . Calculating the partial derivatives: V x = x (6xy - y + 2yz) = 6y V y = y (6xy - y + 2yz) = 6x - 1 + 2z V z = z (6xy - y + 2yz) = 2y At the point (1, 0, 1) : V x = 6(0) = 0 V y = 6(1) - 1 + 2(1) = 6 - 1 + 2 = 7 V z = 2(0) = 0 Substituting these values into the expression for E : E = -(0 i + 7 j + 0 k ) = -7 j . Answer: -7 j