By what factor is the magnitude of the electric field at P multiplied when the arc is collapsed to a point at distance R from P?

Steps to Solve

1. Electric Field from Half-Circle Rod

The electric field due to a uniformly charged half-circle rod of radius $R$ and total charge $+Q$ at point $P$ can be calculated using the formula:

$$ E_{arc} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{R^2} $$

2. Electric Field from a Point Charge

When the arc collapses into a point charge located at a distance $R$ from point $P$, the electric field at $P$ due to this point charge becomes:

$$ E_{point} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{(R - R)^2} $$

This expression appears undefined because $R - R = 0$, indicating that as the rod collapses into a point, the nature of the electric field changes drastically.

3. Understanding the Change in Electric Field Magnitude

To find the multiplicative factor of the electric field's magnitude at $P$, we can express it relative to the previous electric field from the half-circle. The point charge's field is infinite as we come closer to the charge. However, we look at the relative changes:

Let’s denote the original field at $P$ as $E_{arc}$ and the new field from the point charge as approaching infinity:

$$ \text{Multiplicative Factor} = \frac{E_{point}}{E_{arc}} $$

Since the electric field of a point charge diverges, we conclude that the factor is effectively infinite as $R \to 0$.