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Question: Field in certain region of space is given as: $\vec{E} = \frac{x\hat{i} + y\hat{j} + z\hat{k}}{x^2 ...

Field in certain region of space is given as:

E⃗=xi^+yj^+zk^x2+y2+z2\vec{E} = \frac{x\hat{i} + y\hat{j} + z\hat{k}}{x^2 + y^2 + z^2}

Compute charge density in space, also calculate flux through a sphere of radius R centered at the origin.

Answer

Charge Density: ρ=ϵ0r2\displaystyle \rho = \frac{\epsilon_0}{r^2}

Flux through a sphere of radius RR: ΦE=4πR\displaystyle \Phi_E = 4\pi R

Explanation

Solution

Solution

  1. Finding the Charge Density

The electric field is given by

E⃗=xi^+yj^+zk^x2+y2+z2.\vec{E} = \frac{x\hat{i} + y\hat{j} + z\hat{k}}{x^2+y^2+z^2}.

In spherical coordinates, where r=x2+y2+z2r = \sqrt{x^2+y^2+z^2} and r^=xi^+yj^+zk^r\hat{r} = \frac{x\hat{i} + y\hat{j} + z\hat{k}}{r}, the field becomes

E⃗=rr^r2=r^r.\vec{E} = \frac{r\hat{r}}{r^2} = \frac{\hat{r}}{r}.

For a radial field E⃗=f(r)r^\vec{E} = f(r)\hat{r}, the divergence in spherical coordinates is

∇⋅E⃗=1r2ddr[r2f(r)].\nabla \cdot \vec{E} = \frac{1}{r^2}\frac{d}{dr}\Big[r^2 f(r)\Big].

Here, f(r)=1rf(r)=\frac{1}{r} so that

r2f(r)=r2⋅1r=r.r^2 f(r) = r^2 \cdot \frac{1}{r} = r.

Differentiating,

ddr(r)=1.\frac{d}{dr}(r) = 1.

Thus,

∇⋅E⃗=1r2.\nabla \cdot \vec{E} = \frac{1}{r^2}.

By Gauss's law in differential form,

∇⋅E⃗=ρϵ0,\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0},

hence the charge density is

ρ=ϵ0 ∇⋅E⃗=ϵ0r2.\rho = \epsilon_0 \, \nabla \cdot \vec{E} = \frac{\epsilon_0}{r^2}.
  1. Calculating the Flux Through a Sphere of Radius RR

The total electric flux through a closed surface is

ΦE=∮E⃗⋅dA⃗.\Phi_E=\oint \vec{E}\cdot d\vec{A}.

For a sphere of radius RR, the outward area element is

dA⃗=r^ R2sin⁡θ dθ dϕ.d\vec{A} = \hat{r}\,R^2\sin\theta\,d\theta\,d\phi.

Since E⃗=r^R\vec{E} = \frac{\hat{r}}{R} on the surface of the sphere, the dot product becomes

E⃗⋅dA⃗=1RR2sin⁡θ dθ dϕ=R sin⁡θ dθ dϕ.\vec{E}\cdot d\vec{A} = \frac{1}{R} R^2 \sin\theta\,d\theta\,d\phi = R\,\sin\theta\,d\theta\,d\phi.

Integrating over the entire sphere:

ΦE=R∫02πdϕ∫0πsin⁡θ dθ=R⋅(2π)⋅(2)=4πR.\Phi_E = R \int_{0}^{2\pi}d\phi \int_{0}^{\pi}\sin\theta\,d\theta = R \cdot (2\pi) \cdot (2) = 4\pi R.

Explanation (Minimal Core Steps):

  • Rewrite the field in spherical coordinates: E⃗=r^/r\vec{E}=\hat{r}/r.
  • Use ∇⋅E⃗=1r2ddr(r2⋅(1/r))=1r2\nabla\cdot\vec{E}=\frac{1}{r^2}\frac{d}{dr}(r^2 \cdot (1/r))=\frac{1}{r^2}.
  • Apply Gauss’s law: ρ=ϵ0(1/r2)\rho=\epsilon_0 (1/r^2).
  • Flux through a sphere: Φ=E(R)⋅4πR2=1R⋅4πR2=4πR\Phi=E(R)\cdot 4\pi R^2=\frac{1}{R}\cdot4\pi R^2=4\pi R.