Two infinitely long thin straight wires AB and CD with charge densities $-\dfrac{12\,\mu\text{C}}{\text{m}}$ and $+\dfrac{12\,\mu\text{C}}{\text{m}}$ are lying parallel to the y-axis at points $(-1\ \text{m}, 0, 0)$ and $(1\ \text{m}, 0, 0)$ respectively, as shown in the figure. Find the net electric field $\vec{E}$ at points $P_1$ $(-2\ \text{m}, 0, 0)$, O $(0, 0, 0)$ and $P_2$ $(2\ \text{m}, 0, 0)$.
Show Hint
Use \(E = 2k\lambda/r\) for each wire. The field points away from the positive wire and toward the negative wire. Add the two along the x-axis.
Step 1: Understanding the Concept:
An infinitely long line charge of linear density \(\lambda\) produces a field that points radially away from the wire if \(\lambda\) is positive. It points toward the wire if \(\lambda\) is negative.
The field is perpendicular to the wire, so here it lies along the x-axis. The wires are parallel to the y-axis.
Step 2: Key Formula or Approach:
The field at perpendicular distance \(r\) from a long line charge is:
\[ E = \frac{\lambda}{2\pi \varepsilon_0 r} = \frac{2k\lambda}{r} \]
Here \(k = 9 \times 10^{9}\ \text{N m}^2\text{C}^{-2}\). The net field is the vector sum of the fields of the two wires.
Step 3: Find the basic value of the field:
Here \(\lambda = 12\ \mu\text{C/m} = 12 \times 10^{-6}\ \text{C/m}\).
\[ 2k\lambda = 2 \times 9 \times 10^{9} \times 12 \times 10^{-6} = 2.16 \times 10^{5}\ \text{N m C}^{-1} \]
So \(E = 2.16 \times 10^{5}/r\) in N/C when \(r\) is in metres.
Step 4: Field at \(P_1\) (-2 m, 0, 0):
Wire AB (negative) is at x = -1 m, so \(r = 1\) m. The field points toward the wire, along \(+\hat{i}\).
\[ E_{AB} = \frac{2.16 \times 10^{5}}{1} = 2.16 \times 10^{5}\ \text{N/C}\ (+\hat{i}) \]
Wire CD (positive) is at x = +1 m, so \(r = 3\) m. The field points away from the wire, along \(-\hat{i}\).
\[ E_{CD} = \frac{2.16 \times 10^{5}}{3} = 0.72 \times 10^{5}\ \text{N/C}\ (-\hat{i}) \]
\[ \vec{E}_{P_1} = (2.16 - 0.72) \times 10^{5}\,\hat{i} = 1.44 \times 10^{5}\,\hat{i}\ \text{N/C} \]
Step 5: Field at O (0, 0, 0):
O is 1 m from each wire. The negative wire AB is on the left, so its field at O points toward it, along \(-\hat{i}\).
The positive wire CD is on the right, so its field at O points away from it, also along \(-\hat{i}\). The two fields add.
\[ \vec{E}_O = -(2.16 + 2.16) \times 10^{5}\,\hat{i} = -4.32 \times 10^{5}\,\hat{i}\ \text{N/C} \]
Step 6: Field at \(P_2\) (2 m, 0, 0):
Wire CD is 1 m away. Its field points away from it, along \(+\hat{i}\), with magnitude \(2.16 \times 10^{5}\) N/C.
Wire AB is 3 m away. Its field points toward it, along \(-\hat{i}\), with magnitude \(0.72 \times 10^{5}\) N/C.
\[ \vec{E}_{P_2} = (2.16 - 0.72) \times 10^{5}\,\hat{i} = 1.44 \times 10^{5}\,\hat{i}\ \text{N/C} \]
Final Answer:
The net fields are \(1.44 \times 10^{5}\,\hat{i}\) N/C at \(P_1\), \(-4.32 \times 10^{5}\,\hat{i}\) N/C at O, and \(1.44 \times 10^{5}\,\hat{i}\) N/C at \(P_2\).
\[ \boxed{\vec{E}_{P_1} = 1.44 \times 10^{5}\,\hat{i},\ \vec{E}_O = -4.32 \times 10^{5}\,\hat{i},\ \vec{E}_{P_2} = 1.44 \times 10^{5}\,\hat{i}\ \text{N/C}} \]
