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 22:40, 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} $