Question

Write down the electric field due to a point charge in cartesian co-ordinates and hence show that the curl of this field is zero.

Answer 1

needs to be in cartesian co-ordinates

Answer 2

\[r=\sqrt{x^2+y^2+z^2}\]\[\mathbf r = x\,\hat{\mathbf x}+y\,\hat{\mathbf y}+z\,\hat{\mathbf z}\]

Answer 3

\[\begin{align}\mathbf E &= K\frac{Q}{r^2}\hat{\mathbf r}\\
&=K\frac{Q}{r^3}{\mathbf r}\\
&=K\frac Q{(x^2+y^2+z^2)^{3/2}}(x\,\hat{\mathbf x}+y\,\hat{\mathbf y}+z\,\hat{\mathbf z})
\end{align}\]

Answer 4

\[
E_x = \frac{Qx}{(x^2+y^2+z^2)^{3/2}}\\
E_y = \frac{Qy}{(x^2+y^2+z^2)^{3/2}}\\
E_z = \frac{Qz}{(x^2+y^2+z^2)^{3/2}}
\]

Answer 5

\[\nabla \times\mathbf E = \begin{vmatrix}\hat{\mathbf x}&\hat{\mathbf y}&\hat{\mathbf z}\\\frac\partial{\partial x}&\frac\partial{\partial y}&\frac\partial{\partial z}\\E_x&E_y&E_z \end{vmatrix}\\
\qquad \quad=\left(\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}\right)\hat{\mathbf x}
+\left(\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x}\right)\hat{\mathbf y}
+\left(\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}\right)\hat{\mathbf x}\]

Answer 6

\[E_x = \frac{Qx}{(x^2+y^2+z^2)^{3/2}}\]
\[\frac{\partial E_x}{\partial y} = \]

Answer 7

i think i have got it , but the working is horrible

Answer 8

Is there a simpler method?

Answer 9

so we have to prove that the field is irrotational right?

Answer 10

we can assume the field
\[E=E_x \hat{x}+E_y \hat{y}+E_z \hat{z}\]

Answer 11

and corresponding potential field
\[V=V_x \hat{x}+V_y \hat{y}+V_z \hat{z}\]

Answer 12

now we can use the relation
\[E=- \nabla V\]

Answer 13

it will produce
\[\nabla \times E=- \nabla \times (\nabla V)\]

Answer 14

makes sense?

Answer 15

now using this relation -