Using Maxwell's equation in vacuum, derive the wave equation for the y component of the electric field vector
Answers
Explanation:
\overrightarrow{\nabla}\times\overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t }
∇
×
E
=−
∂t
∂
B
\overrightarrow{\nabla}\times\overrightarrow{B}=\mu_0\epsilon_0\frac{\partial \overrightarrow{E}}{\partial t }
∇
×
B
=μ
0
ϵ
0
∂t
∂
E
\overrightarrow{\nabla}\times\overrightarrow{E}(z,t)\overrightarrow{i}=\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ \frac{\partial }{\partial x } & \frac{\partial }{\partial y } & \frac{\partial }{\partial z } \\ \overrightarrow{E}(z,t) &0 & 0 \end{vmatrix}=\frac{\partial E}{\partial z }\overrightarrow{j}
∇
×
E
(z,t)
i
=
∣
∣
∣
∣
∣
∣
∣
∣
i
∂x
∂
E
(z,t)
j
∂y
∂
0
k
∂z
∂
0
∣
∣
∣
∣
∣
∣
∣
∣
=
∂z
∂E
j
\frac{\partial E}{\partial z }=-\frac{\partial B}{\partial t }
∂z
∂E
=−
∂t
∂B
\overrightarrow{\nabla}\times\overrightarrow{B}(z,t)\overrightarrow{j}=\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ \frac{\partial }{\partial x } & \frac{\partial }{\partial y } & \frac{\partial }{\partial z } \\ 0 & \overrightarrow{B}(z,t) & 0 \end{vmatrix}=-\frac{\partial B}{\partial z }\overrightarrow{i}
∇
×
B
(z,t)
j
=
∣
∣
∣
∣
∣
∣
∣
∣
i
∂x
∂
0
j
∂y
∂
B
(z,t)
k
∂z
∂
0
∣
∣
∣
∣
∣
∣
∣
∣
=−
∂z
∂B
i
\frac{\partial B}{\partial z }=-\mu_0\epsilon_0\frac{\partial E}{\partial t }
∂z
∂B
=−μ
0
ϵ
0
∂t
∂E
\frac{\partial^2 E}{\partial z^2 }=-\frac{\partial}{\partial z }\frac{\partial B}{\partial t }=-\frac{\partial}{\partial t }\frac{\partial B}{\partial z }=-\frac{\partial}{\partial t }(-\mu_0\epsilon_0\frac{\partial E}{\partial t })=\mu_0\epsilon_0\frac{\partial^2 E}{\partial t^2 }
∂z
2
∂
2
E
=−
∂z
∂
∂t
∂B
=−
∂t
∂
∂z
∂B
=−
∂t
∂
(−μ
0
ϵ
0
∂t
∂E
)=μ
0
ϵ
0
∂t
2
∂
2
E
\frac{\partial^2 E}{\partial z^2 }=\mu_0\epsilon_0\frac{\partial^2 E}{\partial t^2 }
∂z
2
∂
2
E
=μ 0
ϵ
0 ∂t
2 ∂ 2 E Answer