A point charge `q` is situated at a distance `r` from one end of a thin conduction rod of length `L` having a charge `Q` (uniformly distributed a long its length).find the magnitudes of electric force between the two.

Options

A. `(Kqq)/r^(2)`
B. `(2KQ)/(r(r+L))`
C. `(KQq)/(r(r-L))`
D. `(KQq)/(r(r+L))`

Answer

To find the electric force between a point charge \( q \) situated at a distance \( r \) from one end of a thin conducting rod of length \( L \) with a uniformly distributed charge \( Q \), we can follow these steps:

### Step 1: Define Charge Density
The charge per unit length \( \lambda \) of the rod is given by:
\[ \lambda = \frac{Q}{L} \]

### Step 2: Consider an Element of the Rod
Take an infinitesimal element of the rod at a distance \( x \) from the point charge \( q \) with thickness \( dx \). The charge \( dq \) on this element is:
\[ dq = \lambda \, dx = \frac{Q}{L} \, dx \]

### Step 3: Calculate the Electric Field Due to \( dq \)
The electric field \( dE \) at the location of charge \( q \) due to the charge \( dq \) is given by Coulomb's law:
\[ dE = k \frac{dq}{x^2} = k \frac{\frac{Q}{L} \, dx}{x^2} \]
where \( k \) is Coulomb's constant.

### Step 4: Set Up the Integral for the Total Electric Field
The total electric field \( E \) at the location of charge \( q \) due to the entire rod can be found by integrating \( dE \) from \( x = r \) to \( x = r + L \):
\[ E = \int_{r}^{r+L} dE = \int_{r}^{r+L} k \frac{\frac{Q}{L} \, dx}{x^2} \]

### Step 5: Perform the Integration
The integral can be simplified as follows:
\[ E = k \frac{Q}{L} \int_{r}^{r+L} \frac{dx}{x^2} \]
The integral \( \int \frac{dx}{x^2} \) evaluates to \( -\frac{1}{x} \):
\[ E = k \frac{Q}{L} \left[-\frac{1}{x}\right]_{r}^{r+L} = k \frac{Q}{L} \left(-\frac{1}{r+L} + \frac{1}{r}\right) \]
\[ E = k \frac{Q}{L} \left(\frac{1}{r} - \frac{1}{r+L}\right) \]

### Step 6: Calculate the Force on Charge \( q \)
The force \( F \) on the charge \( q \) due to the electric field \( E \) is given by:
\[ F = qE \]
Substituting the expression for \( E \):
\[ F = q \cdot k \frac{Q}{L} \left(\frac{1}{r} - \frac{1}{r+L}\right) \]

### Step 7: Simplify the Expression
Now, simplifying the expression for force:
\[ F = k \frac{qQ}{L} \left(\frac{(r+L) - r}{r(r+L)}\right) = k \frac{qQ}{L} \frac{L}{r(r+L)} \]
Thus, the final expression for the electric force \( F \) is:
\[ F = k \frac{qQ}{r(r+L)} \]

### Final Result
The magnitude of the electric force between the point charge \( q \) and the conducting rod with charge \( Q \) is:
\[ F = k \frac{qQ}{r(r+L)} \]