Answer:
The energy in the spark produced by discharging the second capacitor is more energetic than the discharge spark of the first capacitor.
Explanation:
Given that the two isolated parallel plate have the same plate area A and
are given the same charge Q.
Energy stored in the capacitor [tex]= \frac{Q^2}{2C}[/tex]
[tex]E = \frac{Q^2}{2C}\\\\Q = \sqrt{2EC} \\\\\sqrt{2E_1C_1} = \sqrt{2E_2C_2}\\\\\sqrt{E_1C_1} = \sqrt{E_2C_2}[/tex]
Where;
C is the capacitance and it is given as;
[tex]C =\frac{\epsilon A}{d}[/tex]
where;
ε is a constant known as permittivity of free space
substitute C in the above equation;
[tex]\sqrt{E_1*\frac{A}{d}} = \sqrt{E_2*\frac{A}{D}}\\\\\sqrt{\frac{E_1}{d}} = \sqrt{\frac{E_2}{D}} \\\\E_2 =\frac{D*E_1}{d}, \ D >d, \ thus \ E_2>E_1[/tex]
Therefore, the energy in the spark produced by discharging the second capacitor is more energetic than the discharge spark of the first capacitor.
