The electric potential at points in an xy plane is given by V = (2.0 V/m2)x2 -(3.5 V/m2)y2. What are (a) the magnitude of the electric field at the point (4.1 m, 2.8 m) and (b) the angle that the field there makes with the positive x direction.

The electric potential at points in an xy plane is given by V = (2.0 V/m²)x² - (3.5 V/m²)y². What are:

(a) the magnitude of the electric field at the point (4.1 m, 2.8 m) and;

(b) the angle that the field there makes with the positive x direction.

(a) 25.56 V/m

(b) 129.92°

Write the electric potential in vector form as follows;

V = (2.0)x² i  - (3.5)y² j              ------------------(i)

Where;

i and j are unit vectors in the x and y direction respectively

The electric field (E) is given as the negative derivative of electric potential (V) with respect to x and y as follows;

E = - derivative [V]

E = - ([2 * 2.0x] i - [2 * 3.5 y] j )

E = - (4.0x i - 7.0y j )

E = - 4.0x i + 7.0y j        -----------------------------(ii)

Now to get the electric field at point (4.1m, 2.8m), substitute the values of x = 4.1m and y = 2.8m into equation (ii) as follows;

E = -4.0(4.1) i + 7.0(2.8) j

E = - 16.4i + 19.6j        ---------------------(iii)

The magnitude of E ( | E |) is calculated as follows;

| E | =

| E | =

| E | =

| E | = 25.56V/m

Therefore, the magnitude of the electric field is 25.56V/m

(b) The direction, θ, of the electric field is given as

tan θ = (-19.6 / 16.4)

tan θ = - 1.195

θ = tan⁻¹ (-1.195)

θ = -50.08°

The negative sign shows that it is measured with respect to the -x axis.

To get the equivalent as measured with respect to the +ve axis, note that from equation (iii), it is evident that the electric field is at -i and +j counterclockwise from the +x-axis.

Therefore, add 180° to -50.08° to measure the direction counterclockwise from the positive x direction as follows;

true angle = 180°  - 50.08° = 129.92°

Therefore the angle that the field makes with the positive x axis is 129.92°

32x+ 21y

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