Question

A uniform spherical volume charge distribution of radius \(2\) m, centered at the origin, has a strength of \(\frac{3}{\pi}\times10^{-6}\) C/m\(^3\). A point charge of strength \(\pi\times8.854\times10^{-12}\) C is moved from \((-3,0,-4)\) to \((0,0,4)\) in Cartesian coordinate system. The relative permittivity of the medium is \(1\) and the coordinate values are in meters.
The work done during the process is \(\mu\)J (round off to two decimal places).

Hint:

Since both points lie outside the charged sphere, treat it as a point charge equal to the total enclosed charge.

Answer 1

Correct Answer: 0.4

Step 1: Get the total charge on the sphere.
The charge density is $\rho=\dfrac{3}{\pi}\times10^{-6}$ C/m$^3$ over a sphere of radius $2$ m, so the total charge is
$Q=\rho\cdot\dfrac{4}{3}\pi(2)^3=\dfrac{3}{\pi}\times10^{-6}\times\dfrac{32\pi}{3}=32\times10^{-6}$ C.

Step 2: Check both endpoints sit outside the sphere.
$(-3,0,-4)$ is at distance $r_1=\sqrt{9+16}=5$ m from the origin, and $(0,0,4)$ is at distance $r_2=4$ m. Since the sphere only extends to $2$ m, both points are in the region outside the charge, where the sphere behaves exactly like a point charge $Q$ at the origin.

Step 3: Write down Coulomb's constant.
With relative permittivity $1$, use $k=\dfrac{1}{4\pi\varepsilon_0}\approx8.988\times10^9$ N$\cdot$m$^2$/C$^2$. The potential at a distance $r$ from the sphere (outside it) is $V(r)=\dfrac{kQ}{r}$.

Step 4: Compute both potentials numerically.
$V_1=\dfrac{kQ}{r_1}=\dfrac{8.988\times10^9\times32\times10^{-6}}{5}\approx57520$ V.
$V_2=\dfrac{kQ}{r_2}=\dfrac{8.988\times10^9\times32\times10^{-6}}{4}\approx71900$ V.

Step 5: Find the potential difference.
$\Delta V=V_2-V_1\approx71900-57520=14380$ V.

Step 6: Bring in the moving charge and multiply.
The moving charge is $q=\pi\times8.854\times10^{-12}\approx2.7818\times10^{-11}$ C. The work done is $W=q\,\Delta V$.
$W\approx2.7818\times10^{-11}\times14380\approx4.00\times10^{-7}$ J.

Step 7: Convert to microjoules.
$4.00\times10^{-7}$ J is the same as $0.40\ \mu$J.
\[\boxed{0.40\ \mu\text{J}}\]