| Line 8: | Line 8: | ||
<math> | <math> | ||
| − | \nabla V = \ | + | \nabla V = \left[\begin{array}{l} |
| − | + | \frac{\partial}{\partial x}(x^3yz+2y^2z+xz^4) \\ | |
| + | \frac{\partial}{\partial y}(x^3yz+2y^2z+xz^4) \\ | ||
| + | \frac{\partial}{\partial z}(x^3yz+2y^2z+xz^4) | ||
| + | \end{array}\right] \\ | ||
| + | \nabla V = | ||
| + | \left[\begin{array}{l} | ||
| + | 3x^2yz + z^4 \\ | ||
| + | x^3z + 4yz \\ | ||
| + | x^3y + 2y^2 + 4xz^3 | ||
| + | \end{array}\right] | ||
</math> | </math> | ||
| + | |||
| + | Applying the identity, we get: | ||
| + | |||
| + | <math> | ||
| + | E = | ||
| + | \left[\begin{array}{l} | ||
| + | -3x^2yz - z^4 \\ | ||
| + | -x^3z - 4yz \\ | ||
| + | -x^3y - 2y^2 - 4xz^3 | ||
| + | \end{array}\right] | ||
| + | </math> | ||
| + | |||
| + | [[File:Vector field|thumbnail|How this electric field looks in 3D space]] | ||
b. The charge density of the field at <math>(2,-3)</math> | b. The charge density of the field at <math>(2,-3)</math> | ||
Revision as of 21:40, 6 December 2020
Electric Potential Sample Problem
Given electric potential equation $ V = x^3yz+2y^2z+xz^4 $, find:
a. The corresponding electric field equation for this potential
Using the identity $ E = - \nabla V $, we know that we need to compute the gradient of $ V $. We get:
$ \nabla V = \left[\begin{array}{l} \frac{\partial}{\partial x}(x^3yz+2y^2z+xz^4) \\ \frac{\partial}{\partial y}(x^3yz+2y^2z+xz^4) \\ \frac{\partial}{\partial z}(x^3yz+2y^2z+xz^4) \end{array}\right] \\ \nabla V = \left[\begin{array}{l} 3x^2yz + z^4 \\ x^3z + 4yz \\ x^3y + 2y^2 + 4xz^3 \end{array}\right] $
Applying the identity, we get:
$ E = \left[\begin{array}{l} -3x^2yz - z^4 \\ -x^3z - 4yz \\ -x^3y - 2y^2 - 4xz^3 \end{array}\right] $
b. The charge density of the field at $ (2,-3) $
$ E = - \nabla V $
$ \nabla \cdot E = \frac{\rho}{\epsilon_0} $
$ \Delta V = -\Large\frac{\rho}{\epsilon_0} $
