Step 1: Recall standard ellipse formulas.
For ellipse $ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 $ with $ a > b $: latus rectum length $ = \dfrac{2b^2}{a} $, minor axis length $ = 2b $. Eccentricity $ e = \sqrt{1 - \dfrac{b^2}{a^2}} $.
Step 2: Set up the given condition.
Latus rectum = half of minor axis means: \[ \frac{2b^2}{a} = \frac{1}{2}(2b) = b \]
Step 3: Simplify to find $ b $ in terms of $ a $.
From $ \dfrac{2b^2}{a} = b $: multiply both sides by $ a $: $ 2b^2 = ab $. Divide by $ b $ ($ b \neq 0 $): $ 2b = a $, so $ b = \dfrac{a}{2} $.
Step 4: Substitute into the eccentricity formula.
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{(a/2)^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]
Step 5: Verify the result is a valid eccentricity.
For an ellipse, $ 0 < e < 1 $. $ \dfrac{\sqrt{3}}{2} \approx 0.866 $, which satisfies this constraint.
Step 6: State the answer.
\[ \boxed{\frac{\sqrt{3}}{2}} \]