0
$\begingroup$

As we know the electric field at a point due to charge $q$ is equal to $\frac{kq}{r^2}$ where $k$ is the constant of proportionality and $r$ is the distance between the charge and the point (where we are finding the electric field). But, in the case of a charged spherical sphere (having charge $q$ and radius $R$), the electric field on the surface of the sphere is $\frac{kq}{R^2}$. I could not understand as to why, in the books, they are taking the distance $R$. The charge is on the surface of the sphere, not at the center of the sphere.

$\endgroup$
  • $\begingroup$ HINT:-Use Gauss's law $\endgroup$ – Unique May 18 at 8:48
0
$\begingroup$

As we know the electric field at a point due to charge $q$ is equal to $\frac{kq}{r^2}$.

Yes, but one has to realize that this result is for a point charge $q$. You see, for instance, only then you can talk about the distance between the point at which you are calculating the field and the charge. If the charge were not a point charge, what would be the distance? There is no well-defined notion of the distance of a point from an extended object.

But, in the case of a charged spherical sphere (having charge $q$ and radius $R$), the electric field on the surface of the sphere is $\frac{kq}{R^2}$. I could not understand as to why, in the books, they are taking the distance $R$. The charge is on the surface of the sphere, not at the center of the sphere.

Yes, you are partially on the right track, if they were directly applying the formula $\frac{kq}{r^2}$ then to put $r=R$, the charge $q$ had to be at the center. But, the point is, we cannot directly use the formula $\frac{kq}{r^2}$ as the charge $q$ is distributed across the entire shell and is not situated at a single point. The formula is only valid if the entire charge $q$ were at any single point. Think about it, even if you try to apply the formula directly to calculate the field on the surface of the shell, what distance would you take as $r$? You cannot take $r=0$ because some amount of charge is certainly at a non-zero distance from any given point on the surface. You cannot take any particular distance because there is always some part of the charge that is situated at a different distance from any given point on the surface. So, the wisdom is in realizing that you cannot use the formula $\frac{kq}{r^2}$ directly as the charge $q$ is simply not situated at any single point.


Now, what do you do? Well, you realize that you can use the familiar formula but only for each of the infinitesimally small charges $dq$ situated in the infinitesimal vicinity of every point on the shell. And thus, you use the formula $\frac{kdq}{r^2} \hat{r}$ to calculate the electric field created by the infinitesimal charge distribution $dq$ and then sum over all such infinitesimal charges which are distributed over the shell. In other words, you perform the integration $\int\frac{kdq}{r^2}\hat{r}$. This will give you the quoted result $\frac{kq}{R^2}$ if you meticulously perform the integration. Or, you can use the symmetry arguments to convince yourself that the electric field must point radially outwards at each point on the spherical shell and then use the Gauss's law to find the same result much more easily.

$\endgroup$
-2
$\begingroup$

enter image description hereenter image description here

This is what my book has to say about this. To answer your question directly, the electric field at any point inside a charged hollow sphere is zero, regardless of what point on a Gaussian surface you take for calculation. For better understanding, i suggest you read Gauss Law and its applications

$\endgroup$
  • $\begingroup$ Please don't upload images as answers. This answer also addresses a different question to the OP which asks for the field outside a sphere, not inside. $\endgroup$ – jacob1729 May 18 at 9:09
  • $\begingroup$ -1: In general, if you don't have something relatively customized to post as an answer, consider leaving a comment. Anyway, as @jacob1729 pointed out, this is not really addressing the question that OP asked either. $\endgroup$ – Feynmans Out for Grumpy Cat May 18 at 9:27

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.