Question

Consider a sphere of radius 5 mm placed in free space containing a total amount of charge ‘Q’. The electric potential due to the distribution of the charge is V = 7 R2 Sin θ Cos ϕ. Find out the numeric value of the charge Q enclosed by the surface.

Answer 1

Let the potential be:

$$V = 7r^2 \sin\theta\cos \varphi$$

According to the Maxwell's equations, the charge density will be:

$$\bigtriangleup V = -\dfrac{\rho}{\varepsilon_0}$$

where $\rho$ is the charge density and $\varepsilon_0$ is the dielectric constant.

The Laplace operator in the spherical coordinates will have the following view:

$$\Delta V = \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial V}{\partial r} \right) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left(\sin \theta \frac{\partial V}{\partial \theta} \right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 V}{\partial \varphi^2}$$

The first term:

$$\frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial V}{\partial r} \right) = \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \cdot14r\sin\theta\cos \varphi \right) = \\ =\frac{1}{r^2}\cdot42r^2\sin\theta\cos \varphi =42\sin\theta\cos \varphi$$

The second term:

$$\frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left(\sin \theta \frac{\partial V}{\partial \theta} \right) =\frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left(\sin \theta \cdot 7r^2 \cos\theta\cos \varphi \right) =\\ =\frac{7\cos \varphi}{2 \sin \theta} \frac{\partial}{\partial \theta} \left(\sin2\theta \right) = \frac{7\cos \varphi\cos\theta}{\sin \theta} =7\cos\varphi\tan\theta$$

The third term:

$$\frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 V}{\partial \varphi^2} = \frac{7r^2\sin\theta}{r^2 \sin^2 \theta} \frac{\partial^2 \cos\varphi}{\partial \varphi^2} =\frac{7}{\sin \theta} \frac{\partial^2 \cos\varphi}{\partial \varphi^2} = -\frac{7\cos\varphi}{\sin \theta}$$

Putting it all together:

$$\rho = 7\varepsilon_0 \left( \frac{\cos\varphi}{\sin \theta}- 6\sin\theta\cos \varphi - \cos\varphi\tan\theta \right)$$

How to find the total charge, one should integrate this density over whole sphere:

$$Q = \iiint\rho dV = \iiint\rho r^2 \sin\theta drd\theta d\varphi$$

The integration w.r. to $r$ will give the multiplier $r_0^3/3$ , where $r_0 = 5 mm$ is the radius of the sphere, because the charge density does not depend on the radial coordinate.

Integrating the first term of density:

$$\iint\frac{\cos\varphi}{\sin \theta} \sin\theta d\theta d\varphi = 4\int^\pi _0d\theta\int^{\pi/2}_0 \cos\varphi d\varphi = 4\pi\sin\varphi|^{\pi/2}_0 = 4\pi$$

Integrating the second term of density:

$$-\iint6\sin^2\theta\cos \varphi d\theta d\varphi =-6 \cdot4\int^\pi _0 \sin^2\theta d\theta \int^{\pi/2}_0 \cos\varphi d\varphi = -6\cdot\dfrac{\pi}{2}\cdot 4\pi = -48\pi^2$$

Integrating the third term of density:

$$-\iint \tan\theta\cos \varphi \sin\theta d\theta d\varphi =- \int^\pi _0 \tan\theta\sin\theta d\theta \int^{2\pi}_0 \cos\varphi d\varphi = 0$$

Finally:

$$Q = 7\varepsilon_0 \dfrac{r_0^3}{3} \left(4\pi - 48\pi^2 \right) = 7\cdot8.85\cdot 10^{-12} \dfrac{0.005^3}{3} \left(4\pi - 48\pi^2 \right) = -1.2\cdot 10^{-15} C$$

Answer. Q = -1.2*10^(-15) C