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 | = [tex]\sqrt{(-16.4)^2 + (19.6)^2}[/tex]
| E | = [tex]\sqrt{(268.96 + 384.16)}[/tex]
| E | = [tex]\sqrt{(653.12)}[/tex]
| 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°
