When n identical capacitors are connected in series across a battery, the energy stored is u. if the same capacitors were connected in parallel across the same battery, then how much energy would be stored?
nu
n2u
u/n
u

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Below is the solution:

capacitors are connected in series 

same q same V1=V/N

U1=1/2 C*(V/N)^2

U(total-1)=N*U1=C*V^2/2N

capacitors connected in parallel

same V

U2=1/2 C*V^2

U(total-2)=1/2 N*C*V^2

U(total-2)/U(total-1)= C*V^2/2N/N*C*V^2/2=>U(total-2)=N^2* U(total-1)

there are no components of a physical goal mentioned on the list of choices that you've provided.