Line 8: Line 8:
  
 
<math>
 
<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 = \left[\begin{array}{l}
\nabla V = \Big[3x^2yz + z^4,x^3z + 4yz,x^3y + 2y^2 + 4xz^3\Big]
+
\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] $

File:Vector field
How this electric field looks in 3D space

b. The charge density of the field at $ (2,-3) $


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$ E = - \nabla V $

$ \nabla \cdot E = \frac{\rho}{\epsilon_0} $


$ \Delta V = -\Large\frac{\rho}{\epsilon_0} $

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Dhruv Lamba, BSEE2010