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(The hint from the book is that, charge is conserved.)

Here is how I have approached the problem:

Q= CV, so the charge on [tex]C_1[/tex] is 7.7 * 10^-6 * 125 = 9.625 * 10^-4

And this is also the total charge before removing the battery. After inserting [tex]C_2[/tex], the total voltage is 30V, so we have:

[tex] 9.625*10^{-4} = (C_1 + C_2)(30)[/tex]

and solve for [tex] C_2[/tex], which is 2.4 * 10^-5

Am i right?