When the decimal point of a certain positive decimal number is moved four places to the right, the new number is four times the reciprocal of the original number. What is the original number?

$ \textbf{(A) }0.0002\qquad\textbf{(B) }0.002\qquad\textbf{(C) }0.02\qquad\textbf{(D) }0.2\qquad\textbf{(E) }2$
Let the original number be \( x \). According to the problem, when the decimal point of \( x \) is moved four places to the right, the new number is \( 4 \times \frac{1}{x} \).

1. Moving the decimal point four places to the right is equivalent to multiplying the number by \( 10^4 \). Therefore, the new number is \( 10^4 x \).

2. According to the problem, this new number is also equal to \( 4 \times \frac{1}{x} \). Therefore, we can set up the equation:
   \[
   10^4 x = 4 \times \frac{1}{x}
   \]

3. To solve for \( x \), first multiply both sides of the equation by \( x \) to eliminate the fraction:
   \[
   10^4 x^2 = 4
   \]

4. Next, divide both sides by \( 10^4 \):
   \[
   x^2 = \frac{4}{10^4} = \frac{4}{10000} = 0.0004
   \]

5. Take the square root of both sides to solve for \( x \):
   \[
   x = \sqrt{0.0004} = 0.02
   \]

Thus, the original number is \( 0.02 \).

The final answer is \( \boxed{0.02} \).