Answer:
1/2
Step-by-step explanation:
The energy stored in a capacitor is given by
![U=(1)/(2)CV^2](https://img.qammunity.org/2020/formulas/physics/high-school/mfr8da5e4v0v1o5i84auoik8ibualt75si.png)
where
C is the capacitance
V is the potential difference
Calling
the capacitance of capacitor 1 and
its potential difference, the energy stored in capacitor 1 is
![U=(1)/(2)C_1 V_1^2](https://img.qammunity.org/2020/formulas/physics/high-school/cgabg2kxod8ry3vaol8zfl67y8sm9kwc52.png)
For capacitor 2, we have:
- The capacitance is half that of capacitor 1:
![C_2 = (C_1)/(2)](https://img.qammunity.org/2020/formulas/physics/high-school/frh77by76b65t07oa6a3bk9wlivzq1wjqc.png)
- The voltage is twice the voltage of capacitor 1:
![V_2 = 2 V_1](https://img.qammunity.org/2020/formulas/physics/high-school/el4bi3utiyyj8rvr0yioweob06o91cthk9.png)
so the energy stored in capacitor 2 is
![U_2 = (1)/(2)C_2 V_2^2 = (1)/(2)(C_1)/(2)(2V_1)^2 = C_1 V_1^2](https://img.qammunity.org/2020/formulas/physics/high-school/tw6ilyi2m0fc9n009przmv6cu5vk6vxkix.png)
So the ratio between the two energies is
![(U_1)/(U_2)=((1)/(2)C_1 V_1^2)/(C_1 V_1^2)=(1)/(2)](https://img.qammunity.org/2020/formulas/physics/high-school/69y3qy0zdoirveoq1dikkuh6q3100js7a0.png)