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Using Lenz's Law

When inner tw...

Question

When inner two concentric spheres of radius $r_{1}$ and $r_{2}$ ( $r_{1} < r_{2}$ ) carries an electric charge ,the differential equation for the potential $V$ at the distance $r$ from the common centre is $\frac{d^{2} V}{d r^{2}} + \frac{2}{r} = 0$ . The value of $V$ in terms of $r$ is?

A

V r = c_{1} r - c_{2}

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B

V r^{2} = c_{1} r - c_{2}

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C

V r = c_{1} r^{2} - c_{2}

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D

V r = c_{1} r + c_{2}

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Solution

The correct option is C $V r = c_{1} r - c_{2}$
Given Differential equation is $\frac{d^{2} V}{d r^{2}} = - \frac{2}{r} \frac{d V}{d r}$ ,

Let $\frac{d v}{d r} = u$ ,

Now the equation becomes,

$\frac{d u}{d r} = - \frac{2}{r} u$ ,

Separating the variables and integrating gives,

$u = \frac{c_{1}}{r^{2}}$ ,

Substitutung the value of $u$ gives,

$\frac{d V}{d r} = \frac{c_{1}}{r^{2}}$ ,

Separating the variables and integrating gives,

$V r = c_{1} r \pm c_{2}$ ,

Where $c_{1}, c_{2}$ both are constants.

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