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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: |
| − | b. The charge density of the field at <math>(2,-3)</math> | + | <math> |
| + | \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> | ||
| + | |||
| + | 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> | ||
| + | |||
| + | <center> | ||
| + | [[File:Electric field representation.jpeg|701px|thumbnail|center|How this electric field looks like in 3D space]] | ||
| + | </center> | ||
| + | |||
| + | b. The charge density of the field at <math>(-2,3,1)</math> | ||
| + | |||
| + | Using the identity <math> | ||
| + | \nabla \cdot E = \frac{\rho}{\epsilon_0} | ||
| + | </math>, we know we need to take the divergence of the electric field. We get: | ||
| + | |||
| + | <math>\nabla \cdot E = | ||
| + | \left[\begin{array}{l} | ||
| + | \frac{\partial}{\partial x} \\ | ||
| + | \frac{\partial}{\partial y} \\ | ||
| + | \frac{\partial}{\partial z} | ||
| + | \end{array}\right] | ||
| + | \cdot | ||
| + | \left[\begin{array}{l} | ||
| + | -3x^2yz - z^4 \\ | ||
| + | -x^3z - 4yz \\ | ||
| + | -x^3y - 2y^2 - 4xz^3 | ||
| + | \end{array}\right] \\ | ||
| + | |||
| + | \nabla \cdot E = -6xyz - 4z - 12xz^2 | ||
| + | </math> | ||
| + | |||
| + | Noting that <math>E = - \nabla V</math>, you should realize that by taking the divergence of the electric field, we have computed the Laplacian of the voltage equation. While you must always technically compute the equation for an electric field by taking the Laplacian of the voltage, the important thing to note is that the Laplacian itself creates a concrete relationship between voltage and charge density. Changing notation, we get: | ||
| + | |||
| + | <math> | ||
| + | \nabla \cdot E = \Delta V = -6xyz - 4z - 12xz^2 \\ | ||
| + | \frac{\rho}{\epsilon_0} = -6xyz - 4z - 12xz^2 \\ | ||
| + | \rho = \epsilon_0 * (-6xyz - 4z - 12xz^2)\\ | ||
| + | </math> | ||
| + | |||
| + | Plugging in our coordinates, we get: | ||
| + | |||
| + | <math> | ||
| + | \rho = \epsilon_0 * (-6*(-2)*3*1 - 4*1 - 12*(-2)*1^2) \\ | ||
| + | \rho = 56\epsilon_0 | ||
| + | </math> | ||
| + | |||
| + | Coulombs per cubic meter, assuming SI units | ||
| − | |||
[[Walther_MA271_Fall2020_topic9|Back to main page]] | [[Walther_MA271_Fall2020_topic9|Back to main page]] | ||
Latest revision as of 22:17, 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 $ (-2,3,1) $
Using the identity $ \nabla \cdot E = \frac{\rho}{\epsilon_0} $, we know we need to take the divergence of the electric field. We get:
$ \nabla \cdot E = \left[\begin{array}{l} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{array}\right] \cdot \left[\begin{array}{l} -3x^2yz - z^4 \\ -x^3z - 4yz \\ -x^3y - 2y^2 - 4xz^3 \end{array}\right] \\ \nabla \cdot E = -6xyz - 4z - 12xz^2 $
Noting that $ E = - \nabla V $, you should realize that by taking the divergence of the electric field, we have computed the Laplacian of the voltage equation. While you must always technically compute the equation for an electric field by taking the Laplacian of the voltage, the important thing to note is that the Laplacian itself creates a concrete relationship between voltage and charge density. Changing notation, we get:
$ \nabla \cdot E = \Delta V = -6xyz - 4z - 12xz^2 \\ \frac{\rho}{\epsilon_0} = -6xyz - 4z - 12xz^2 \\ \rho = \epsilon_0 * (-6xyz - 4z - 12xz^2)\\ $
Plugging in our coordinates, we get:
$ \rho = \epsilon_0 * (-6*(-2)*3*1 - 4*1 - 12*(-2)*1^2) \\ \rho = 56\epsilon_0 $
Coulombs per cubic meter, assuming SI units
