Why is this equation from Purcell's EM textbook correct?

[Leo Liu]

Homework Statement

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Relevant Equations

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I saw the following equation on page 31 in Purcell's EM textbook.
$$F=\epsilon_0\int_{E_1}^{E_2} E\, dE=\frac 2 {\epsilon_0} (E_2^2-E_1^2)$$
Here, F is the force on a unit area.
And then he claims that since $E_2-E_1=\sigma/\epsilon_0$, the equation can be further simplified to
$$F=\frac 1 {\epsilon_0}(E_1+E_2)\sigma$$

However, I think the correct coefficient in the last part of first equation should be the inverse of what it is now ($\frac{\epsilon_0}{2}$ instead), as only then can I obtain the second expression. I have no idea why the author wrote the equation this way. To give it a little bit of background, dE is the change in the electric field of a thin layer of a charged sphere, E1 is the the electric field inside, and E2 is the electric field outside. Could someone explain?

 

[Delta2]

yes, at least viewing from a purely mathematical view (integrating $\int x dx=\frac{x^2}{2}$ that should have been $\frac{\epsilon_0}{2}$. Also from a dimensional analysis point of view $\epsilon_0$ just cannot be on the denominator, the units won't match then.

 

[Steve4Physics]

Homework Statement:: .
Relevant Equations:: .

I saw the following equation on page 31 in Purcell's EM textbook.
$$F=\epsilon_0\int_{E_1}^{E_2} E\, dE=\frac 2 {\epsilon_0} (E_2^2-E_1^2)$$

The formula (1.48 in the book) is in fact
$\frac F A = … = \frac {\epsilon_0}{2}(E_2^2 – E_1^2)$
Check your book!

 

[Leo Liu]

The formula (1.48 in the book) is in fact
$\frac F A = … = \frac {\epsilon_0}{2}(E_2^2 – E_1^2)$
Check your book!

Here is a picture of my book. This book is probably an older edition which is authorized to be printed in China by the original publisher.

 

[Steve4Physics]

;View attachment 287228
Here is a picture of my book. This book is probably an older edition which is authorized to be printed in China by the original publisher.

I guess this is an error in earlier editions which has now been corrected. The 3rd edition (bottom of page 31) shows this: