Question

The electric potential at any point (x,y) is given by V(x,y)=xy+x²+y² Volts. Find out the electric field at (1 m,2 m).

Answer 1

The electric field at the point (1 m, 2 m) when the electric potential is given by V(x,y) = xy + x² + y² Volts is calculated to be –→E = (-4 V/m, -5 V/m).

Step-by-step explanation:

The electric potential at any point (x,y) is given by V(x,y) = xy + x² + y² Volts. To find the electric field at the point (1 m, 2 m), we need to calculate the negative gradient of the potential function. The electric field components can be found by taking the partial derivatives of the potential with respect to x and y respectively:

Ex = -∂V/∂x = -(y + 2x)

Ey = -∂V/∂y = -(x + 2y)

Substituting x = 1m and y = 2m, we get:

Ex(1,2) = -(2 + 2(1)) = -4 V/m

Ey(1,2) = -(1 + 2(2)) = -5 V/m

Therefore, the electric field at the point (1 m, 2 m) is –→E = (Ex, Ey) = (-4 V/m, -5 V/m).