Question E5D10:
As a conductor’s diameter increases, what is the effect on its electrical length?  

Choices:  
A: Thickness has no effect on electrical length  
B: It varies randomly  
C: It decreases  
D: It increases    

Correct Answer is  D: It increases  

As a conductor (antenna element) gets thicker, its fringing capacitance to space increases (“end effect”). That extra capacitance makes a given physical length behave as if it were longer electrically, so it resonates at a lower frequency unless you shorten it. Therefore, electrical length increases.

Why D (It increases) is correct:
- Increased diameter ⇒ increased end-effect capacitance ⇒ lower resonant frequency for the same physical length ⇒ the element is electrically longer.

Why the other choices are incorrect:
- A: Thickness has no effect — Incorrect. Diameter clearly affects capacitance, bandwidth, and resonant length.
- B: It varies randomly — Incorrect. The effect is systematic: thicker generally means electrically longer.
- C: It decreases — Incorrect. The opposite occurs; you typically must shorten a fat element to keep the same resonant frequency.

Useful formulas and rules of thumb:
- Electrical length in radians:  
  $$\theta = \beta \, l = \frac{2\pi}{\lambda}\, l$$  
  If the element’s diameter increases, effective electrical length θ increases for the same physical length l.
- Half-wave dipole (thin wire, free space) length:  
  $$L_{\text{thin}} \text{(ft)} \approx \frac{468}{f_{\text{MHz}}}$$
- With thicker or insulated wire, use a shortening factor $k<1$:  
  $$L_{\text{fat}} \approx k \cdot \frac{468}{f_{\text{MHz}}}, \quad k \approx 0.90\text{–}0.98$$  
  Smaller $k$ implies more electrical length (you cut the element shorter).
- Rule of thumb: Thicker or insulated elements are electrically longer, resonate lower, and have wider bandwidth.