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a. The corresponding electric field equation for this potential | a. The corresponding electric field equation for this potential | ||
| − | Using the identity | + | Using the identity <math>E = - \nabla V</math>, we know that we need to compute the gradient of <math>V</math>. We get: |
| + | |||
| + | <math> | ||
| + | \nabla V = \Bigg[\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)\Bigg]\\ | ||
| + | \nabla V = \Big[3x^2yz + z^4,x^3z + 4yz,x^3y + 2y^2 + 4xz^3\Big] | ||
| + | |||
| + | </math> | ||
b. The charge density of the field at <math>(2,-3)</math> | b. The charge density of the field at <math>(2,-3)</math> | ||
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<math></math> | <math></math> | ||
[[Walther_MA271_Fall2020_topic9|Back to main page]] | [[Walther_MA271_Fall2020_topic9|Back to main page]] | ||
| + | |||
| + | <math>E = - \nabla V</math> | ||
| + | |||
| + | <math> | ||
| + | \nabla \cdot E = \frac{\rho}{\epsilon_0} | ||
| + | </math> | ||
| + | |||
| + | |||
| + | <math> | ||
| + | \Delta V = -\Large\frac{\rho}{\epsilon_0} | ||
| + | </math> | ||
Revision as of 22:22, 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 = \Bigg[\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)\Bigg]\\ \nabla V = \Big[3x^2yz + z^4,x^3z + 4yz,x^3y + 2y^2 + 4xz^3\Big] $
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} $
