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 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.

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    $\begingroup$ HINT:-Use Gauss's law $\endgroup$ May 18, 2019 at 8:48

4 Answers 4


You are right that only point charges satisfy $\vec{E} = kq\hat{r}/r^2$ exactly. In general, given a random charge distribution $\rho(\vec{r})$, you have to do an integral to find the electric field on each point (Or equivalently, solve the Poisson equation, but let's not talk about that for the moment).

However, your book is equally right in claiming that the electric field on a sphere is given by $\vec{E}=kq\hat{r}/R^2$. In fact, for a uniformly charges sphere, the $\vec{E}$ field outside the sphere is exactly equal to the $\vec{E}$ field that would be been produced if all the charges are concentrated at the center of the sphere.

Why is it so? We can look at this question in two ways:

  1. Using shell theorem Shell Theorem (Wikipedia)
  2. Using Gauss law Gauss law (Wikipedia)

Shell theorem

For the first approach (Shell theorem), we do things by first principles. Since the relevant Wikipedia page has all the details and the computation is tedious, I will only walk through the concept:

  1. First, we assume the charge density depends only on $r$ (i.e. $\rho(r)$)
  2. Then, to calculate the total effect of a whole charged sphere, we only need to calculate that from from each shell, and sum them (integrate) all the effect of all these shells to get the final results
  3. As it turns out, the field produced by a uniformly charged shell of radius $R$ with charge $q$ is simple: \begin{align*} \vec{E}_\text{shell} = \begin{cases} \dfrac{kq\hat{r}}{r^2} &\quad \text{for} \quad r > R\\ \vec{0} &\quad \text{for} \quad r < R \end{cases} \end{align*}
  4. Now break up the sphere into shells and add up their effect. It should not be hard to see that the resulting field outside the sphere is exactly the same as if all the charges are concentrated at the center.

Gauss law

There is also another (more elegant) way to look at this using Gauss law. By Coulomb's law: \begin{align*} \vec{E} = \frac{1}{4\pi \epsilon_0} \frac{q}{r^2} \hat{r} \end{align*} where $k=1/4\pi \epsilon_0$. Now, consider an arbitrary closed surface $S$ enclosing the charge, and calculate the electric flux passing through the surface.

\begin{align*} \Phi_E = \iint_S \vec{E} \cdot \hat{n} dS \end{align*}

What does this equation means? It means you (mentally) cut the surface $S$ into small patches, find the outward pointing unit normal vector for each patch, and calculate the E-fields that "flows out" of the surface by $\vec{E} \cdot \hat{n} dS$, then you sum it over the whole surface.

Now, this integral can be evaluated directly using solid angles:

\begin{align*} \Phi_E &= \iint_S \vec{E} \cdot \hat{n} dS\\ &= \iint_S \frac{1}{4\pi \epsilon_0} \frac{q}{r^2} \hat{r} \cdot \hat{n} dS\\ &= \int_\Omega \frac{1}{4\pi \epsilon_0} \frac{q}{r^2} r^2d\Omega\\ &= \frac{q}{\epsilon_0} \frac{1}{4\pi} \int_\Omega d\Omega\\ &= \frac{q}{\epsilon_0} \end{align*}

So the electric flux passing through an arbitrary surface is always $q_\text{enclosed}/\epsilon_0$. In fact, direct application of superposition principle shows that this hold for any charge distribution. So we have:

\begin{align*} \iint_\text{closed} \vec{E} \cdot \hat{n} dS = \frac{1}{\epsilon_0}\sum_i Q^\text{enclosed}_i \end{align*}

Applying this law to find the field for sphere is easy.

  1. The sphere is spherically symmetric, so by symmetry concerns $\vec{E}(r)$ is a function of radius about the center only. Furthermore it must pointing radially outward (i.e. $\vec{E} = E(r) \hat{r}$).
  2. Construct a close spherical surface $S$ of radius $r$ enclosing the charged sphere and calculate the flux passing through it. It should be easy to see that: \begin{align*} \Phi_E = 4\pi r^2 E(r) \end{align*}
  3. Now since our spherical surface enclose the charged sphere, the enclosed charge is the total charge of the sphere $q$. Hence: \begin{align*} 4\pi r^2 E(r) &= q/\epsilon_0\\ E(r) &= \frac{1}{4\pi \epsilon_0} \frac{q}{r^2} \end{align*}

Substituting $r=R$ in the result should recover the equation your book mentioned.


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.


It is not difficult to show that the flux of electric field going out from a point charge through a closed surface of any shape or size depends only on the size (and sign) of the charge (inside of the surface). Given that, then a spherically symmetric charge centered inside a spherical surface will produce the same flux (and field) as that produced by a point charge of the same size located at the center of the sphere. (Simple applications of Gauss's law generally require a high degree of symmetry.)


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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

  • 1
    $\begingroup$ 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, 2019 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. $\endgroup$
    – user87745
    May 18, 2019 at 9:27
  • $\begingroup$ Don't post pictures of other reference materials as answers. $\endgroup$
    – Yejus
    Jul 26, 2021 at 14:53

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