To find the electric potential at point \((x, 0)\) due to a uniformly charged rod, we need to integrate the potential contributions from each infinitesimal charge element along the rod. The rod extends along the X-axis from \(0\) to \(l\) with total charge \(Q\).
First, let's define an infinitesimal element of charge \(dq\) on the rod located at position \(x'\) along the X-axis. We can express \(dq\) as:
\(dq = \lambda \, dx'\), where \(\lambda = \frac{Q}{l}\) is the linear charge density.
The potential \(dV\) at point \((x, 0)\) due to \(dq\) is given by:
\(dV = \frac{k \, dq}{x - x'}\), where \(k = \frac{1}{4 \pi \epsilon_0}\).
Substitute \(dq\) in the formula:
\(dV = \frac{1}{4 \pi \epsilon_0} \frac{\lambda \, dx'}{x - x'}\).
Integrate \(dV\) from \(0\) to \(l\):
\(V(x) = \int_0^l \frac{\lambda \, dx'}{x - x'} = \frac{Q}{4 \pi \epsilon_0 l} \int_0^l \frac{dx'}{x - x'}\).
This integral evaluates to:
\(V(x) = \frac{Q}{4 \pi \epsilon_0 l} \left[ -\log_e(x - x') \right]_0^l\).
On evaluating the limits, we get:
\(V(x) = \frac{Q}{4 \pi \epsilon_0 l} \left[ -\log_e(x - l) + \log_e(x) \right]\)
\(= \frac{Q}{4 \pi \epsilon_0 l} \log_e\left(\frac{x}{x - l}\right)\).
Thus, the potential at point \((x, 0)\) when \(x > l\) is:
\(V(x) = \frac{Q}{4 \pi \epsilon_0 l} \log_e\left(\frac{x}{x - l}\right)\).
Thus, the correct answer is:
\(\frac{Q}{4 \pi \epsilon_0 l} \log_e \left(\frac{x}{x - l}\right)\).
A 10 $\mu\text{C}$ charge is placed in an electric field of $ 5 \times 10^3 \text{N/C} $. What is the force experienced by the charge?