2
$\begingroup$

When we derive the formula for energy of a continuous charge distribution $\rho$ using this equation $$W = \frac{1}{2}\int\rho V \text d\tau$$ with $V$ being the electric potential, we get this expression for work done $$W = \frac{\epsilon}{2}\left[\int E^2\text d\tau + \int V\mathbf E\cdot \text d\mathbf a\right]$$ where $\mathbf E$ is the electric field.

Now in the book it is mentioned that the contribution from term $\int V\mathbf E\cdot\text d\mathbf a$ will approach to zero as we make our Gaussian surface larger and larger because $EV$ will vary as $\frac{1}{r^3}$ and multiplying it with by $r^2$ roughly make it $\frac{1}{r}$. By the same logic then shouldn't the contribution from the term $\int E^2d\tau$ approach to zero? But its written that this whole integral is positive since $E^2$ will only increase

$\endgroup$
3
$\begingroup$

No. The volume integral is over all space contained in the surface, so using a larger volume just adds to the integral. You aren't just looking at $E^2$ at the boundary of the surface. In other words, you are looking at all $r$ values, not just where the boundary is.

Contrast this with the surface integral where you actually are just looking at everything at the boundary of the surface. At this point you can think of how the values drop off or grow with $r$ because you're only looking at at single $r$ value when evaluating the integral.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.