Question

show that the function V= A/r+B satisfies Laplace's equation, where A abd B are constant and r is the magnitude of position vector

Answer 1

here is your answer.

Answer 2

Answer:

The function satisfy the Laplace equation $\sqrt{\frac{\partial^{2} v}{\partial r^{2}}}+\frac{2}{r} \frac{\partial v}{\partial r}=0$

Explanation:

Laplace's equation serves this purpose. Even if we don't know V, it's highly likely that we know anything about its divergence. The electrostatic potential is represented by Laplace's equation, which follows directly from Gauss and Faraday's principles.

$\begin{aligned}&\frac{d v}{d r}=-\frac{A}{r^{2}}+0 \leq-\frac{A}{r^{2}} \\&\frac{d^{2} v}{d i^{2}}=+ \frac{2A}{r^{3}} \\&\frac{d^{2} v}{d v^{2}}=-\frac{2}{r} \frac{d v}{d r} \\&\Rightarrow \frac{d^{2} v}{d r^{2}}+\frac{2}{r} \frac{d v}{d r} \infty 0\end{aligned}$

Laplace's equation in spherical coordinates -

if $v$ depends only on $r$, then $\frac{\partial v}{\partial \theta}=0$

$\Rightarrow$equation becomes $-\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{-} \frac{\partial v}{\partial r}\right)=0$

or,

$r^{2}-\frac{\partial^{2} v}{\partial r^{2}}+2 r \frac{\partial v}{\partial r}=0$

$\Rightarrow \frac{\partial^{2} v}{\partial r^{2}}+\frac{2}{r} \frac{\partial v}{\partial r}=0$

The function satisfy the Laplace equation $\sqrt{\frac{\partial^{2} v}{\partial r^{2}}}+\frac{2}{r} \frac{\partial v}{\partial r}=0$