Answer to Question #87257 in Electricity and Magnetism for SAURABH

Question #87257
Using Maxwell’s equations in free space, derive the wave equation for the z-component of
the electric field vector.
1
Expert's answer
2019-04-02T10:48:52-0400

In free space we have the following relation between the vectors (in Gaussian system of units):


"\\vec{D}=\\vec{E}\\\\\n\\vec{B}=\\vec{H}"

Hence, the Maxwell's equations transform to the form as follows:


"\\nabla \\cdot \\vec{E} = \\rho\\\\\n \\nabla \\times \\vec{E} = - \\frac{1}{c} \\frac{\\partial \\vec{H}}\n{\\partial t}\\\\\n\\nabla \\cdot \\vec{H} = 0\\\\ \n\\nabla \\times \\vec{H} = \\frac{1}{c} \\frac{\\partial \\vec{E}}{\\partial t}\t+ \\frac{4 \\pi}{c} \\vec{j}"

Calculating the curl from both the left and right side of the 2nd equation we derive:


"\\nabla \\times (\\nabla \\times \\vec{E}) = \\nabla(\\nabla \\cdot \\vec{E}) - \\Delta \\vec{E} = \\nabla \\rho - \\Delta \\vec{E} \\\\\n\\nabla \\times \\left( - \\frac{1}{c} \\frac{\\partial \\vec{H}}{\\partial t} \\right) = - \\frac{1}{c} \\frac{\\partial}{\\partial t} \\left( \\frac{1}{c} \\frac{\\partial \\vec{E}}{\\partial t}\t\n \t+ \\frac{4 \\pi}{c} \\vec{j} \\right) = - \\frac{1}{c^2} \\frac{\\partial^2 \\vec{E}}{\\partial t^2} - \\frac{4 \\pi}{c^2} \\frac{\\partial \\vec{j}}{\\partial t}\\\\\n\\Rightarrow \\, \\Delta \\vec{E} - \\frac{1}{c^2} \\frac{\\partial^2 \\vec{E}}{\\partial t^2} = \\nabla \\rho + \\frac{4 \\pi}{c^2} \\frac{\\partial \\vec{j}}{\\partial t}"

Finally, projecting this vector equation on the z-axis, we obtain:


"\\Delta E_z - \\frac{1}{c^2} \\frac{\\partial^2 E_z}{\\partial t^2} = \\frac{\\partial \\rho}{\\partial z} + \\frac{4 \\pi}{c^2} \\frac{\\partial j_z}{\\partial t}"


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