A capacitor is made with two square plates, each with side length $L$, of negligible thickness, and capacitance $C$. The two-plate capacitor is put in a microwave which increases the side length of each square plate by $ 1 \% $. By what percent does the voltage between the two plates in the capacitor change? 
	
	$ \textbf {(A) } \text {decreases by } 2\% \\ \textbf {(B) } \text {decreases by } 1\% \\ \textbf {(C) } \text {it does not change} \\ \textbf {(D) } \text {increases by } 1\% \\ \textbf {(E) } \text {increases by } 2\% $

[i]Problem proposed by Ahaan Rungta[/i]
1. **Understanding the problem**: We are given a capacitor with square plates of side length \( L \) and capacitance \( C \). The side length of each plate increases by \( 1\% \) when placed in a microwave. We need to determine the percentage change in the voltage between the plates.

2. **Capacitance of a parallel plate capacitor**: The capacitance \( C \) of a parallel plate capacitor is given by:
   \[
   C = \frac{\epsilon_0 A}{d}
   \]
   where \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of one of the plates, and \( d \) is the separation between the plates.

3. **Area of the plates**: Since the plates are square with side length \( L \), the area \( A \) is:
   \[
   A = L^2
   \]

4. **New side length after increase**: The side length increases by \( 1\% \), so the new side length \( L' \) is:
   \[
   L' = L \times 1.01
   \]

5. **New area of the plates**: The new area \( A' \) is:
   \[
   A' = (L')^2 = (L \times 1.01)^2 = L^2 \times (1.01)^2 = L^2 \times 1.0201
   \]

6. **New capacitance**: The new capacitance \( C' \) is:
   \[
   C' = \frac{\epsilon_0 A'}{d} = \frac{\epsilon_0 (L^2 \times 1.0201)}{d} = C \times 1.0201
   \]

7. **Voltage relationship**: The voltage \( V \) across the capacitor is related to the charge \( Q \) and capacitance \( C \) by:
   \[
   V = \frac{Q}{C}
   \]

8. **New voltage**: The new voltage \( V' \) is:
   \[
   V' = \frac{Q}{C'} = \frac{Q}{C \times 1.0201} = \frac{V}{1.0201}
   \]

9. **Percentage change in voltage**: The percentage change in voltage is:
   \[
   \frac{V' - V}{V} \times 100\% = \left( \frac{\frac{V}{1.0201} - V}{V} \right) \times 100\% = \left( \frac{1}{1.0201} - 1 \right) \times 100\%
   \]

10. **Simplifying the percentage change**: 
    \[
    \frac{1}{1.0201} \approx 0.9802
    \]
    \[
    \left( 0.9802 - 1 \right) \times 100\% = -0.0198 \times 100\% = -1.98\%
    \]

11. **Conclusion**: The voltage decreases by approximately \( 2\% \).

The final answer is \( \boxed{ \text{decreases by } 2\% } \)