This problem involves calculating the electric field and force due to a uniformly charged rod and a point charge. Here's the breakdown of the questions:

1. Problem Setup:
   • A uniformly charged rod of length $a$ and total charge $Q$ lies along the $x$-axis from $x = 0$ to $x = a$.
   • A point charge $q$ is located at $x = a + r$.

Solutions:

a) Electric Field at a Point $x$ for $x > a$:

The total electric field at a point $x$ is due to contributions from all charge elements of the rod. Let the linear charge density be: $\lambda = \frac{Q}{a}.$ Consider a small charge element $dq = \lambda dx'$ at position $x'$ on the rod. The distance from this element to the point $x$ is $x - x'$. The electric field due to $dq$ at $x$ is: $dE = \frac{1}{4\pi \varepsilon_0} \frac{dq}{(x - x')^2}.$

Substitute $dq = \lambda dx'$: $dE = \frac{1}{4\pi \varepsilon_0} \frac{\lambda dx'}{(x - x')^2}.$

Integrate over the rod ($x'$ runs from 0 to $a$): $E_x = \int_0^a \frac{1}{4\pi \varepsilon_0} \frac{\lambda}{(x - x')^2} dx' = \frac{\lambda}{4\pi \varepsilon_0} \int_0^a \frac{dx'}{(x - x')^2}.$

Performing the integration: $\int_0^a \frac{dx'}{(x - x')^2} = \left[ -\frac{1}{x - x'} \right]_0^a = -\frac{1}{x - a} + \frac{1}{x}.$

Thus: $E_x = \frac{\lambda}{4\pi \varepsilon_0} \left( \frac{1}{x} - \frac{1}{x - a} \right).$

Substitute $\lambda = Q/a$: $E_x = \frac{Q}{4\pi \varepsilon_0 a} \left( \frac{1}{x} - \frac{1}{x - a} \right).$

b) Force on the Point Charge $q$:

The force is given by: $F = qE.$ Using the result for $E_x$ from part (a), substitute $x = a + r$: $E_x = \frac{Q}{4\pi \varepsilon_0 a} \left( \frac{1}{a + r} - \frac{1}{r} \right).$

Thus: $F = q \cdot \frac{Q}{4\pi \varepsilon_0 a} \left( \frac{1}{a + r} - \frac{1}{r} \right).$

Simplify: $F = \frac{qQ}{4\pi \varepsilon_0 a} \left( \frac{r - (a + r)}{r(a + r)} \right) = \frac{qQ}{4\pi \varepsilon_0} \cdot \frac{-a}{ar(a + r)}.$

c) Approximation for $r \gg a$:

When $r \gg a$, $a + r \approx r$. Substituting this into $F$: $F \approx \frac{qQ}{4\pi \varepsilon_0 r^2}.$

This resembles the force between two point charges $q$ and $Q$ separated by a distance $r$, as expected.

Would you like detailed explanations for any of the integrals, assumptions, or simplifications? Let me know! 😊

Related Questions:

1. What is the physical meaning of $r \gg a$ in this context?
2. How would the electric field change if the rod were infinitely long?
3. Could this scenario be solved using a different coordinate system, like polar coordinates?
4. How does the linear charge density $\lambda$ affect the electric field's magnitude?
5. What happens to the electric field if the point charge $q$ is placed closer to the rod?

Tip:

When approximating for $r \gg a$, always verify that higher-order terms (e.g., $a^2$) can be neglected!