The electric potential in a region of space is given by V(x, y, z) = A(xy + yz + zx) . Given that A is 10 SI units, determine the magnitude of the electric field at the point (1 m , 1 m , 1 m ) .

Step-by-step solution

The electric field E is related to the electric potential V by the negative gradient of the potential: E = - ( V x i + V y j + V z k ) Given V(x, y, z) = A(xy + yz + zx) and A = 10 , we have: E_x = - V x = -A(y + z) E_y = - V y = -A(x + z) E_z = - V z = -A(x + y) Substituting x = 1 , y = 1 , z = 1 and A = 10 : E_x = -10(1 + 1) = -20 E_y = -10(1 + 1) = -20 E_z = -10(1 + 1) = -20 The electric field vector is E = -20 i - 20 j - 20 k . The magnitude of the electric field is: | E | = (-20)^2 + (-20)^2 + (-20)^2 = 400 +