(ii) $$ Q_{2} $$ is uniformly distributed along the line charge of length $$ 2 L $$ $$ \rho_{l}=\frac{Q_{2}}{2 L} $$ Electric field $$ \overrightarrow{\mathbf{E}}_{2} $$ at $$ (0,0,0) $$ due to the line charge, $$ \stackrel{\overrightarrow{\mathbf{E}}}{2}=\frac{1}{4 \pi \varepsilon_{0}} \int_{c} \frac{\rho_{l} d l}{R^{2}} \overrightarrow{\mathbf{a}}_{R}=\frac{1}{4 \pi \varepsilon_{0}} \int_{c} \frac{\rho_{l} d l}{R^{3}} \vec{R} $$ Source Point $$ \vec{s}=-a \overrightarrow{\mathbf{a}}_{x}+y \overrightarrow{\mathbf{a}}_{y} $$ Field Point $$ \vec{f}=0, \mathrm{dl}=\mathrm{dy} $$ $$ \overrightarrow{\boldsymbol{R}}=\vec{f}-\vec{s}=a \overrightarrow{\mathbf{a}}_{x}-y \overrightarrow{\mathbf{a}}_{y} $$ $$ \begin{aligned} \mathbf{R} & =\left(a^{2}+y^{2}\right)^{\frac{1}{2}} \ \overrightarrow{\mathbf{E}}_{2} & =\frac{\rho_{l}}{4 \pi \varepsilon_{0}} \int_{-L}^{L} \frac{\left(a \overrightarrow{\mathbf{a}}_{x}-y \overrightarrow{\mathbf{a}}_{y}\right) d y}{\left(a^{2}+y^{2}\right)^{\frac{3}{2}}} \ & =\frac{\boldsymbol{p}_{l}}{4 \pi \varepsilon_{0}} \int_{-L}^{L} \frac{a \overrightarrow{\mathbf{a}}_{x} d y}{\left(a^{2}+y^{2}\right)^{\frac{3}{2}}}(\text { from symmetry }) \ X & =\frac{\rho_{l} a \overrightarrow{\mathbf{a}}_{x}}{4 \pi \varepsilon_{0}} \int_{-L}^{L} \frac{d y}{\left(a^{2}+y^{2}\right)^{\frac{3}{2}}} \end{aligned} $$ SUSS 10

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To determine the electric field $$ \overrightarrow{\mathbf{E}}_{2} $$ at the origin $$(0,0,0)$$ due to a uniformly charged line of length $$ 2L $$ with total charge $$ Q_{2} $$, let's follow the provided steps and complete the integration.

### Given:
- **Linear charge density:**
  $$
  \rho_{l} = \frac{Q_{2}}{2L}
  $$
  
- **Electric field expression:**
  $$
  \overrightarrow{\mathbf{E}}_{2} = \frac{\rho_{l}}{4 \pi \varepsilon_{0}} \int_{-L}^{L} \frac{\left(a \overrightarrow{\mathbf{a}}_{x} - y \overrightarrow{\mathbf{a}}_{y}\right) dy}{\left(a^{2} + y^{2}\right)^{\frac{3}{2}}}
  $$
  
- **Symmetry consideration:** The $$ y $$-components cancel out, leaving only the $$ x $$-component:
  $$
  \overrightarrow{\mathbf{E}}_{2} = \frac{\rho_{l} a \overrightarrow{\mathbf{a}}_{x}}{4 \pi \varepsilon_{0}} \int_{-L}^{L} \frac{dy}{\left(a^{2} + y^{2}\right)^{\frac{3}{2}}}
  $$

### Evaluating the Integral:
Let's compute the integral:
$$
I = \int_{-L}^{L} \frac{dy}{\left(a^{2} + y^{2}\right)^{\frac{3}{2}}}
$$

**Substitution:**
Let $$ y = a \tan \theta $$, hence $$ dy = a \sec^{2} \theta \, d\theta $$.

However, a more straightforward approach uses a standard integral formula:
$$
\int \frac{dy}{\left(a^{2} + y^{2}\right)^{\frac{3}{2}}} = \frac{y}{a^{2} \sqrt{a^{2} + y^{2}}}
$$

**Applying the limits:**
$$
I = \left[ \frac{y}{a^{2} \sqrt{a^{2} + y^{2}}} \right]_{-L}^{L} = \frac{L}{a^{2} \sqrt{a^{2} + L^{2}}} - \left( -\frac{L}{a^{2} \sqrt{a^{2} + L^{2}}} \right) = \frac{2L}{a^{2} \sqrt{a^{2} + L^{2}}}
$$

### Substituting Back into $$ \overrightarrow{\mathbf{E}}_{2} $$:
$$
\overrightarrow{\mathbf{E}}_{2} = \frac{\rho_{l} a \overrightarrow{\mathbf{a}}_{x}}{4 \pi \varepsilon_{0}} \cdot \frac{2L}{a^{2} \sqrt{a^{2} + L^{2}}} = \frac{\rho_{l} \cdot 2L}{4 \pi \varepsilon_{0} \cdot a \sqrt{a^{2} + L^{2}}}} \overrightarrow{\mathbf{a}}_{x}
$$

Simplify the constants:
$$
\overrightarrow{\mathbf{E}}_{2} = \frac{\rho_{l} L}{2 \pi \varepsilon_{0} a \sqrt{a^{2} + L^{2}}}} \overrightarrow{\mathbf{a}}_{x}
$$

### Substituting $$ \rho_{l} $$:
Recall that $$ \rho_{l} = \frac{Q_{2}}{2L} $$:
$$
\overrightarrow{\mathbf{E}}_{2} = \frac{\frac{Q_{2}}{2L} \cdot L}{2 \pi \varepsilon_{0} a \sqrt{a^{2} + L^{2}}}} \overrightarrow{\mathbf{a}}_{x} = \frac{Q_{2}}{4 \pi \varepsilon_{0} a \sqrt{a^{2} + L^{2}}}} \overrightarrow{\mathbf{a}}_{x}
$$

### Final Expression:
$$
\boxed{\overrightarrow{\mathbf{E}}_{2} = \frac{Q_{2}}{4 \pi \varepsilon_{0} \, a \sqrt{a^{2} + L^{2}}} \, \overrightarrow{\mathbf{a}}_{x}}
$$

This expression represents the electric field at the origin due to a uniformly charged line of length $$ 2L $$ and total charge $$ Q_{2} $$. The field points in the $$ x $$-direction and decreases with distance $$ a $$ from the line as well as with increasing $$ L $$.
The electric field $$ \overrightarrow{\mathbf{E}}_{2} $$ at the origin due to a uniformly charged line of length $$ 2L $$ with total charge $$ Q_{2} $$ is: $$ \overrightarrow{\mathbf{E}}_{2} = \frac{Q_{2}}{4 \pi \varepsilon_{0} \, a \sqrt{a^{2} + L^{2}}} \, \overrightarrow{\mathbf{a}}_{x} $$ This field points in the $$ x $$-direction and depends on the distance $$ a $$ from the line and the length $$ L $$ of the charge distribution.

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