Question

Need some algebra help:
Given the equation: 1/C_k=(1/((k*e_0*A)/t)))+(1/(e_0*A)/(d-t))
Solve for k

Answer 1

I tried taking the inverse of both sides and got C_k=e_0*A*k*t+e_0*A(d-t), but I'm not sure if my algebra is correct.

Answer 2

\(\dfrac{1}{C_k}=\dfrac{1}{\dfrac{k*e_0*A}{t}}+(\dfrac{1}{\dfrac{e_0*A}{(d-t)}})\)

\(C_k=\dfrac{e_0*A*k}{t}+\dfrac{e_0*A}{(d-t)}\)

\(C_k-\dfrac{e_0*A}{(d-t)}=\dfrac{e_0*A*k}{t}\)

\(k=[C_k-\dfrac{e_0*A}{(d-t)}]*\dfrac{t}{A*e_0}\)

Answer 3

How would you incorporate \[C_0=\frac{ \epsilon_0*A }{ d }\] into the k= equation?

Answer 4

Can't be incorporated, In the k equation there is no \[C_{0}\]

Answer 5

Supposedly if doing the algebra correctly from the initial given equation, the equation k= should only be in terms of d, t, C_k and C_0. I tried redoing the algebra myself, but I can't find a way to either cancel out the E_0*A or substitute (E_0*A)/d for C_0.

Answer 6

Even from the question I don't see any c_0. It can't come from nowhere suddenly.

Answer 7

C_0 is not directly given from the initial equation, so I assume you had to get the term \[\frac{ \epsilon_0*A }{ d }\] and then substitute C_0 in there, but so far I couldn't figure out a way to manipulate the equation and get that term.