Question

Derive Equation 26.33 for the induced surface charge density on the dielectric in a capacitor.

Short answer

By deriving equation 26.33 for the induced surface charge density on the dielectric in a capacitor we get ${\eta }_i=\left(1-\frac{1}{\kappa }\right){\eta }_0$.

Step-by-step solution

Given Information

Equation 26.33:

${\eta }_{\text{induced}}={\eta }_0\left(1-\frac{1}{\kappa }\right)$

${\eta }_{\text{induced}}$ranges from nearly zero when $\kappa \approx 1$ to $\approx {\eta }_0$ when $\kappa \gg 1$.

Explanation

From expression 26.29 we have

$E_i=E_0-E.$

The induced field will be

$E_i=\frac{{\eta }_i}{{\epsilon }_0}.$

Substitute expression 26.30 to 26.26,

$\kappa :=\frac{E_0}{E}\Rightarrow E_0=\kappa E_0,$

Now find

localid="1648710646958" $E_i=E_0-E=E(\kappa -1)=\left(\frac{\kappa -1}{\kappa }\right)E_0.$

The induced charge density will be related to the induced field as

${\eta }_i=E_i{\epsilon }_0$

Substituting the previous result,

localid="1648710633563" ${\eta }_i=\left(\frac{\kappa -1}{\kappa }\right)E_0{\epsilon }_0$

Note that $E_0{\epsilon }_0={\eta }_0$and rearrange the ratio, to find the result as in the book:

${\eta }_i=\left(1-\frac{1}{\kappa }\right){\eta }_0$

Final Answer

Hence, the expression will be${\eta }_i=\left(1-\frac{1}{\kappa }\right){\eta }_0.$