Question

23.43 • calc in a certain region of space, the electric potential is v(x, y, z)=axy - bx² + cy, where a, b, and c are positive constants. (a) calculate the x-, y-, and z - components of the electric field. (b) at which points is the electric field equal to zero?

Explanation:

Step1: Recall the relationship between electric - field and potential

The electric - field components are given by $E_x=-\frac{\partial V}{\partial x}$, $E_y = -\frac{\partial V}{\partial y}$, and $E_z=-\frac{\partial V}{\partial z}$.

Step2: Calculate $E_x$

Differentiate $V(x,y,z)=Axy - Bx^{2}+Cy$ with respect to $x$:
$E_x=-\frac{\partial V}{\partial x}=-(Ay - 2Bx)=2Bx - Ay$

Step3: Calculate $E_y$

Differentiate $V(x,y,z)$ with respect to $y$:
$E_y=-\frac{\partial V}{\partial y}=-(Ax + C)=-Ax - C$

Step4: Calculate $E_z$

Differentiate $V(x,y,z)$ with respect to $z$:
$E_z=-\frac{\partial V}{\partial z}=0$

Step5: Set $E_x = 0$, $E_y = 0$, and $E_z = 0$ to find the points where $\vec{E}=\vec{0}$

Since $E_z = 0$ always.
Set $E_x=0$: $2Bx - Ay = 0$, so $y=\frac{2B}{A}x$.
Set $E_y=0$: $-Ax - C = 0$, so $x=-\frac{C}{A}$.
Substitute $x = -\frac{C}{A}$ into $y=\frac{2B}{A}x$: $y=\frac{2B}{A}\times(-\frac{C}{A})=-\frac{2BC}{A^{2}}$.

Answer:

(a) $E_x = 2Bx - Ay$, $E_y=-Ax - C$, $E_z = 0$
(b) The electric - field is zero at the point $(x,y,z)=(-\frac{C}{A},-\frac{2BC}{A^{2}},z)$ where $z$ is arbitrary.