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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 \[\hat{r}\]

FAQ (Frequently Asked Questions)

Q1. What is the Electric Field Inside a Conducting Sphere?

Most people often mix up between the electric field inside a conducting sphere with that of an electric field outside a conducting sphere, be it hollow or solid. The electric field inside a sphere is zero, while the electric field outside the sphere can be expressed as: E = kQ/r².

Q2. Why is the Electric Field Inside a Sphere Equal to Zero?

Since the Gauss Law only applies to Gaussian surfaces that do not consume charge and instead, all of its charges lie on the surface. Therefore, following Gaussian symmetry, the electric field inside a sphere is equal to zero. It can be expressed as:

φ = ΦE^{➝}. dA^{➝} = q_{enc}/∊₀

Since the net charge enclosed is zero, therefore, E = 0.