A uniform line charge of \(16 \mathrm{nC} / \mathrm{m}\) is located along the line defined by \(y=\) \(-2, z=5\). If \(\epsilon=\epsilon_{0}:\) (a) find \(\mathbf{E}\) at \(P(1,2,3) .\) (b) find \(\mathbf{E}\) at that point in the \(z=0\) plane where the direction of \(\mathbf{E}\) is given by \((1 / 3) \mathbf{a}_{y}-(2 / 3) \mathbf{a}_{z} .\)

To calculate the electric field due to a line charge at a given point, we use the formula: $$\mathbf{E} = \int_{l} \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R^2}\mathbf{a}_R dl$$ Where \(\lambda\) is the linear charge density, \(R\) is the distance from the line charge to the point of interest, and \(\mathbf{a}_R\) is the unit vector in the radial direction.

We are given point \(P(1, 2, 3)\). The line charge is at \(y=-2\) and \(z=5\). We subtract the coordinates to find the difference in position. $$\Delta y = 2 - (-2) = 4$$ $$\Delta z = 5 - 3 = 2$$ Now we can find the distance and the electric field: $$ R = \sqrt{\Delta y^2 + \Delta z^2} = \sqrt{4^2 + 2^2} = \sqrt{20} = 2\sqrt{5} $$ Now we can find the components of the electric field, \(\mathbf{E}_y\) and \(\mathbf{E}_z\).

The electric field components can be calculated as follows: $$ \mathbf{E}_y = \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R^2}\mathbf{a}_y $$ $$ \mathbf{E}_z = \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R^2}\mathbf{a}_z $$ Now plug in the values we have: $$ \mathbf{E}_y = \frac{9 \times 10^9 \,\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}^2 \cdot 16 \times 10^{-9} \,\mathrm{C}/\mathrm{m}}{(2\sqrt{5})^2} \mathbf{a}_y = 8\sqrt{5} \,\mathrm{N/C} \, \mathbf{a}_y $$ $$ \mathbf{E}_z = \frac{9 \times 10^9 \,\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}^2 \cdot 16 \times 10^{-9} \,\mathrm{C}/\mathrm{m}}{(2\sqrt{5})^2} \mathbf{a}_z = 4\sqrt{5} \,\mathrm{N/C} \, \mathbf{a}_z $$ So, the electric field at point P is: $$\mathbf{E}(1, 2, 3) = 8\sqrt{5} \,\mathrm{N/C} \, \mathbf{a}_y + 4\sqrt{5} \,\mathrm{N/C} \, \mathbf{a}_z$$

We are given the direction of the electric field at a point on the \(z=0\) plane: $$\mathbf{E} = E_y\frac{\mathbf{a}_y}{3} - E_z\frac{2\mathbf{a}_z}{3}$$ The point is on the \(z=0\) plane, so we can write its coordinates as \((x, y, 0)\). The line charge is at \(y=-2\) and \(z=5\). We subtract the coordinates to find the difference in position. $$\Delta y = y - (-2)$$ $$\Delta z = 5 - 0 = 5$$ Now we can find the distance and the electric field: $$ R = \sqrt{\Delta y^2 + \Delta z^2} = \sqrt{(\Delta y)^2 + 5^2} $$ Now we can use the relationship between the electric field components: $$\frac{E_y}{3} = \frac{2E_z}{3}$$ Using the expression for the electric field due to a line charge, we can write: $$\frac{1}{4\pi\epsilon_0} \frac{\lambda}{(\Delta y^2 + 25)^{3/2}}\Delta y = 2 \times \frac{1}{4\pi\epsilon_0} \frac{\lambda}{(\Delta y^2 + 25)^{3/2}} 5$$ Simplifying the equation, we get: $$\Delta y = 10$$ So, the point at which the given electric field direction occurs is \((x,10,0)\). Now we can find the electric field components at this point: $$\mathbf{E}_y = \frac{9 \times 10^9 \,\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}^2 \cdot 16 \times 10^{-9} \,\mathrm{C}/\mathrm{m}}{(5^2 + 10^2)^{3/2}} 10$$ $$\mathbf{E}_z = \frac{9 \times 10^9 \,\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}^2 \cdot 16 \times 10^{-9} \,\mathrm{C}/\mathrm{m}}{(5^2 + 10^2)^{3/2}} 5$$ So, the electric field at the given point on the \(z=0\) plane is: $$\mathbf{E}(x, 10, 0) = \frac{80}{169}\,\mathrm{N/C} \, \mathbf{a}_y + \frac{40}{169}\,\mathrm{N/C} \, \mathbf{a}_z$$