Answer:
Explanation:
plate separation = 2.3 x 10ā»Ā³ m
capacity Cā = Īµ A / d
= Īµ A / 2.3 x 10ā»Ā³
Cā = Īµ A / 1.15 x 10ā»Ā³
[tex]\frac{C_2}{C_1}[/tex] = [tex]\frac{2.3}{1.15}[/tex]
a ) when charge remains constant
energy = [tex]\frac{q^2}{2C}[/tex]
q is charge and C is capacity
energy stored initially Eā= [tex]\frac{q^2}{2C_1}[/tex]
energy stored finally Eā = [tex]\frac{q^2}{2C_2}[/tex]
[tex]\frac{E_1}{E_2} = \frac{C_2}{C_1}[/tex] = [tex]\frac{2.3}{1.15}[/tex]
[tex]E_2[/tex] = [tex]\frac{1.15}{2.3 } \times E_1[/tex]
= [tex]\frac{1.15}{2.3 } \times 8.38[/tex]
= 4.19 J
b )
In this case potential diff remains constant
energy of capacitor = 1/2 C VĀ²
energy is proportional to capacity as V is constant .
[tex]\frac{E_2}{E_1} = \frac{C_2}{C_1}[/tex]
[tex]\frac{E_2}{8.38} = \frac{2.3}{1.15}[/tex]
[tex]E_2[/tex] = 16.76 .
