

Electric Field of a Sphere With Uniform Charge Density
To understand electric fields due to a uniformly charged sphere, first, you need to understand the different types of spherical symmetry. For a uniformly charged conducting sphere, the overall charge density is relative to the distance from the reference point, not on its direction. Therefore, even if you rotate the surface, there would be no difference in its overall charge. Therefore, for a sphere with uniform charge density ρ₀ , is defined as the spherical symmetry, and is represented as:
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If the sphere with radius R is charged to the extent of having the upper half as uniform charge density as ρ₁ while the lower half having a uniform charge density as ρ₂; then it wouldn't have symmetrical charge density because it is variant according to the direction.
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Similarly, for spheres with four distinct shells, the uniform charge density varies, respectively. However, since the charge density function isn't dependent on the direction, rather on its distance, therefore has a symmetry.
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However, for knowing if a surface has spherical symmetry or not, one needs to identify if the charge density depends on the radius r, or on its coordinates ρ(r,θ,φ)
Case I: If the charge density depends on the coordinates ( θ, φ ), then it wouldn't have spherical symmetry
Case II: If the charge density depends on the radius r, it will represent a spherical symmetry.
In general, all spherically symmetrical surfaces must have an electric field radially targeted to any point, and because of its charge, the field needs to be independent of its rotation. Thus, with the spherical coordinates having origins at the center of the spherical charge distribution, the electric field at the point P, with a distance of r from the center can be expressed as:
E\[_{p}^{➝}\] = E\[_{p}\](r) r … eqn. (1)
where, r can be taken as the unit vector that goes from the origin to point P. When radial component E\[_{p}\] > 0, then the vector moves distant from the origin, and when E\[_{p}\] < 0,the vector moves towards the origin.
Electric Field of a Sphere
With the above equation, we can derive the electric field through a Gaussian surface, otherwise known as a closed sphere, with the same center as the center of charge distribution. Therefore, the direction of the area vector of an area element can be represented in the following diagram,
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Here, the radius R expresses the charge distribution while the radius r is the radius of the Gaussian surface. The electric field needs to be directed radially, and the angle between the electric field and area vector is 0 or (cosθ = 1). However, when the angle between the area vector and electric field is 180° or (cosθ = 0).
Therefore, the electric field E\[^{➝}\] has an equal magnitude across all places of the Gaussian surface, sharing the same center for charge distribution. It is represented in the following equation:
Φ\[_{s}\]E\[^{➝}\]ndA = E\[_{p}\]Φ\[_{s}\]dA = E\[_{p}\]4πr²
For cases r> R,
φ = ΦE\[^{➝}\] . dA\[^{➝}\] = \[\frac{q_{enc}}{∊₀}\]
φ = ΦE . dA. cos 0 = \[\frac{σ.4πr²}{σ₀}\]
⇒ EΦdA = \[\frac{σ.4πr²}{σ₀}\]
⇒ E.4πr² = \[\frac{σ.4πr²}{σ₀}\]
E = \[\frac{σ}{∊₀}\] \[\frac{R²}{r²}\]
Thus, the total charge on the sphere is:
q\[_{total}\] = σ.4πr²
The above equation can also be written as:
E = \[\frac{1}{4πr²∊₀}\]\[\frac{q_{total}}{r²}\]
For the net positive charge, the direction of the electric field is from O to P, while for the negative charge, the direction of the electric field is from P to O.
Electric Field Due to Spherical Shell
For a uniformly charged sphere, the charge density that varies with the distance from the centre is:
ρ(r) = arⁿ (r ≤ R; n ≤ 0)
As the given charge density function symbolizes only a radial dependence with no direction dependence, therefore, it can be a spherically symmetrical situation. Therefore, the area of the spherical shell is 4πr’² with a thickness of dr’
Thus, the charge in the shell will be: dq = ar’ⁿ4πr’²dr’
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For electric field due to uniformly charged spherical shell, and at a point outside the charge distribution:
According to Gauss' Law, the radius r > R, the distribution of charge is enclosed within the Gaussian surface. Between r’ = R and r’ = r, it is completely vacant of any charges and thus, expressed as:
q\[_{enc}\] = ∫dq = ∫\[_{0}^{R}\]ar’ⁿ4πr’²dr’ = \[\frac{4πa}{n+3}\] Rⁿ⁺³
Therefore, the electric field E\[^{➝}\] can be expressed as:
E\[_{out}^{➝}\] = [\[\frac{aRⁿ⁺³}{∊₀(n+3)}\]]\[\frac{1}{r²}\]\[\hat{r}\]; where \[\hat{r}\] is the unit vector that goes from the origin to the field point.
Electric Field Inside a Spherical Shell
For a Gaussian surface that has r < R, which is located within the distance r of the center of the spherical charge distribution would be:
q\[_{enc}\] = ∫\[_{0}^{r}\] ar’ⁿ4πr’²dr’ = \[\frac{4πa}{n+3}\] rⁿ⁺³
Therefore, the Electric field, E\[_{int}^{➝}\] can be described as:
E\[_{int}^{➝}\] = [\[\frac{a}{∊₀(n+3)}\]] rⁿ⁺¹ \[\hat{r}\]
FAQs on Electric Field of a Sphere
1. What is the formula for the electric field due to a uniformly charged spherical shell?
The formula for the electric field of a uniformly charged spherical shell (or a hollow sphere) with total charge Q and radius R depends on the distance 'r' from the centre:
- Outside the sphere (r > R): The field is the same as if the entire charge Q were a point charge at the centre. The formula is E = kQ/r², where k = 1/(4πε₀).
- On the surface of the sphere (r = R): The field is at its maximum. The formula is E = kQ/R².
- Inside the sphere (r < R): The electric field is zero. This is because a Gaussian surface drawn inside the shell encloses no charge.
2. Why is the electric field inside any charged hollow sphere or a solid conducting sphere always zero?
The electric field inside a hollow sphere (conducting or non-conducting) or a solid conducting sphere is zero due to the principles of Gauss's Law and electrostatic equilibrium. In a conducting sphere, any excess charge resides entirely on its outer surface. When we consider a Gaussian surface inside the sphere, it encloses no net charge (q_enclosed = 0). According to Gauss's Law (∮E⋅dA = q_enclosed/ε₀), if the enclosed charge is zero, the electric field (E) must also be zero everywhere inside that surface.
3. How is the electric field inside a uniformly charged solid non-conducting sphere different from that of a conducting sphere?
The primary difference lies in the charge distribution and its effect on the internal electric field. In a solid conducting sphere, charge resides only on the surface, making the internal electric field zero. However, in a uniformly charged solid non-conducting sphere, the charge is distributed throughout its entire volume. Therefore, the electric field inside is not zero. It increases linearly with the distance 'r' from the centre, given by the formula E = kQr/R³, where R is the sphere's radius. The field is zero only at the very centre (r=0).
4. What do the electric field lines for a positively charged conducting sphere look like?
The electric field lines for a positively charged conducting sphere have two key characteristics:
- They originate from the surface of the sphere and extend outwards to infinity.
- They are always directed radially outwards and are perpendicular to the surface at every point. This is because the surface of a conductor in electrostatic equilibrium is an equipotential surface.
There are no electric field lines inside the conducting sphere as the field there is zero.
5. How does the electric field of a uniformly charged solid sphere vary with distance from its centre?
The electric field's variation depends on whether the sphere is conducting or non-conducting:
- For a solid conducting sphere: The field is zero from the centre up to the surface. At the surface, it abruptly jumps to its maximum value (kQ/R²) and then decreases with the square of the distance (proportional to 1/r²) outside the sphere.
- For a solid non-conducting sphere: The field starts at zero at the centre and increases linearly up to the surface. It reaches its maximum value (kQ/R²) at the surface and then, similar to the conducting sphere, decreases in proportion to 1/r² outside the sphere.
6. What is a real-world example of the principle that the electric field is zero inside a conducting shell?
A perfect real-world example is a Faraday cage. This is a container made of a conducting material, like a metal mesh. When an external electric field (like from a lightning strike) is applied, the charges on the conductor rearrange themselves to cancel the field's effects inside the cage. This is why being inside a car (which acts as a Faraday cage) during a thunderstorm is relatively safe. It demonstrates the principle of electrostatic shielding, where the interior is protected from external electric fields.
7. Can a non-conducting sphere have a zero electric field inside it?
Yes, but only under specific conditions. A solid non-conducting sphere will have a zero electric field inside only if it contains no net charge. A hollow non-conducting sphere, if uniformly charged on its surface, will also have a zero electric field inside, similar to a hollow conducting sphere. This is because a Gaussian surface drawn inside it would still enclose no charge. However, if a solid non-conducting sphere is uniformly charged throughout its volume, the electric field inside will not be zero, except at its exact centre.

















