A rod of length *L* lies along the *x* axis with its left end at the origin. It has a nonuniform charge density 𝜆 = 𝛼*x*, where 𝛼 is a positive constant.

Calculate the electric potential at point *B* , which lies on the perpendicular bisector of the rod a distance *b* above the *x* axis. (Use the following as necessary: 𝛼, *k*e, *L*, *b*, and *d*.)

To calculate the electric potential \( V_B \) at point \( B \), which lies a distance \( b \) above the perpendicular bisector of the rod, we need to integrate the contributions of the infinitesimal charges along the length of the rod. 

Given that the charge density is non-uniform, \( \lambda = \alpha x \), where \( \alpha \) is a constant and \( x \) is the distance along the rod from the origin, we need to find the potential due to each infinitesimal charge \( dq \).

### Step-by-step breakdown:

1. **Set up the infinitesimal charge element \( dq \):**
   The linear charge density \( \lambda(x) \) varies as:
   \[
   \lambda(x) = \alpha x
   \]
   The infinitesimal charge \( dq \) over an infinitesimal length \( dx \) is given by:
   \[
   dq = \lambda(x) dx = \alpha x dx
   \]

2. **Determine the distance from each charge element to point \( B \):**
   Let the position of point \( B \) be \( (0, b) \). The distance from an infinitesimal charge element at position \( x \) on the rod to the point \( B \) is:
   \[
   d = \sqrt{x^2 + b^2}
   \]

3. **Write the infinitesimal potential \( dV \):**
   The potential at point \( B \) due to an infinitesimal charge \( dq \) is:
   \[
   dV = \frac{k_e dq}{d} = \frac{k_e \alpha x dx}{\sqrt{x^2 + b^2}}
   \]
   where \( k_e \) is the Coulomb constant.

4. **Integrate to find the total potential \( V_B \):**
   The rod extends from \( x = 0 \) to \( x = L \), so we integrate the expression for \( dV \) from 0 to \( L \):
   \[
   V_B = \int_0^L \frac{k_e \alpha x dx}{\sqrt{x^2 + b^2}}
   \]

5. **Solve the integral:**

   This is a standard integral, and the result is:
   \[
   V_B = k_e \alpha \left[ \sqrt{x^2 + b^2} \right]_0^L
   \]

6. **Evaluate the limits:**
   \[
   V_B = k_e \alpha \left( \sqrt{L^2 + b^2} - \sqrt{0^2 + b^2} \right)
   \]
   Simplifying:
   \[
   V_B = k_e \alpha \left( \sqrt{L^2 + b^2} - b \right)
   \]

7. **Substitute \( \alpha \) in terms of total charge (if needed):**
   If we are given the total charge \( Q \) on the rod, we could express \( \alpha \) in terms of \( Q \). However, for this problem, \( \alpha \) is left as a constant, so the final expression for the potential at point \( B \) is:

\[
   V_B = k_e \alpha \left( \sqrt{L^2 + b^2} - b \right)
\]

### Final Answer:
\[
V_B = k_e \alpha \left( \sqrt{L^2 + b^2} - b \right)
\]

Let me know if you want more details or have any questions. Here are five related questions:

1. How does the potential change if the charge density was constant?
2. How would the electric field be calculated from the electric potential in this scenario?
3. How does the potential behave as the distance \( b \) increases significantly?
4. Can you calculate the total charge on the rod if \( \alpha \) is known?
5. How would the potential change if the rod were bent into a semicircle?

**Tip:** When solving for potentials due to non-uniform charge distributions, always consider symmetry and the behavior of charge density carefully.

A rod of length L lies along the x axis with its left end at the origin. It has a nonuniform charge density 𝜆 = 𝛼x, where 𝛼 is a positive constant. Calculate the electric potential at point B, which lies on the perpendicular bisector of the rod a distance b above the x axis. (Use the following as necessary: 𝛼, k_e, L, b, and d.)

Learn how to calculate the electric potential at a point above the perpendicular bisector of a rod with non-uniform charge density (𝜆 = 𝛼x). This solution involves calculus and the superposition principle in electrostatics. Step-by-step integration and key formulas are provided. Suitable for undergraduate physics students.

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