Question

The electric potential in a region of space is V=(210 x^2− 120 y^2)V, where x and y are in meters.

Q1: What is the strength of the electric field at (x,y)=(2.0m,3.0m) ?
Q2: What is the direction of the electric field at (x,y)=(2.0m,3.0m) ? Give the direction as an angle (in degrees) counterclockwise from the positive x-axis.

Answer 1

The strength of the electric field at (2m,3m) is 840 V/m in the x-direction and 720 V/m in the y-direction.

Step-by-step explanation:

To find the strength of the electric field at (2.0m,3.0m), we need to find the gradient of the electric potential function, which gives the electric field. The electric field is the negative gradient of the electric potential. By taking the partial derivatives of the electric potential function with respect to x and y, we can find the components of the electric field.

Given: V = (210x^2 − 120y^2) V

E_x = -dV/dx = -d(210x^2 − 120y^2)/dx = -420x V/m

E_y = -dV/dy = -d(210x^2 − 120y^2)/dy = 240y V/m

Substituting the values (x = 2.0m, y = 3.0m) into the equations, we get:

E_x = -420(2.0) = -840 V/m

E_y = 240(3.0) = 720 V/m

Therefore, the strength of the electric field at (2.0m,3.0m) is 840 V/m in the x-direction and 720 V/m in the y-direction.