Can anyone show me $V(r)=V_0\ln(\frac{b}{a}) \ln(r)-V_0\ln(\frac{b}{a}) \ln(b)=\frac{V_0}{\ln(\frac{a}{b})}\ln(\frac{r}{b})$?

Question

There is an equation, $V(r)=C_1 \ln(r)+C_2$, now I know $V(b)=0$, and $V(a)=V_0$, find $C_1$ and $C_2$.

$V(b)=C_1 \ln(b)+C_2=0$

$V(a)=C_1 \ln(a)+C_2=V_0$

$V_0=C_1(\ln(a)-\ln(b))=C_1\ln{\frac{a}{b}}$, so $C_1=\frac{V_0}{\ln\frac{a}{b}}=V_0\ln(\frac{b}{a})$

$C_2=-C_1 \ln(b)=-V_0\ln(\frac{b}{a}) \ln(b)$ , so $V(r)=V_0\ln(\frac{b}{a}) \ln(r)+-V_0\ln(\frac{b}{a}) \ln(b)$

However, the books says $V(r)=\frac{V_0}{\ln(\frac{a}{b})}\ln(\frac{r}{b})$

Can anyone show me $V(r)=V_0\ln(\frac{b}{a}) \ln(r)-V_0\ln(\frac{b}{a}) \ln(b)=\frac{V_0}{\ln(\frac{a}{b})}\ln(\frac{r}{b})$? Or where am I wrong in my calculation?

Answer 1

"$C_1=\frac{V_0}{ln\frac{a}{b}}=V_0ln(\frac{b}{a})$": no. $\frac1{\ln c}\ne\ln\frac1c.$ ($\ln\frac1c=-\ln c$.)

$$V(r)=C_1\ln r+C_2=C_1(\ln r-\ln b)=\frac{V_0}{\ln\frac ab}\ln\frac rb.$$