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: Find the total charge enclosed by the sphere.
The volume charge density is
\[ \rho=\frac{3}{\pi}\times10^{-6}\ \text{C/m}^3 \]
and the sphere has radius \(R=2\) m. The total charge is
\[ Q=\rho\times\frac{4}{3}\pi R^3=\frac{3}{\pi}\times10^{-6}\times\frac{4}{3}\pi(2)^3 \]
\[ Q=\frac{3}{\pi}\times10^{-6}\times\frac{4}{3}\pi\times8=32\times10^{-6}\ \text{C} \]
The \(\pi\) and the fraction \(3\) cancel neatly, leaving \(Q=32\ \mu\text{C}\).

Step 2: Locate the two points relative to the sphere.
The distance of the point \((-3,0,-4)\) from the origin is
\[ r_1=\sqrt{(-3)^2+0^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5\ \text{m} \]
The distance of the point \((0,0,4)\) from the origin is
\[ r_2=\sqrt{0^2+0^2+4^2}=4\ \text{m} \]
Both \(r_1=5\) m and \(r_2=4\) m are greater than the sphere radius \(R=2\) m, so both points lie outside the charged sphere.

Step 3: Use the point-charge equivalent potential.
Outside a uniformly charged sphere, the electric potential is the same as that of a point charge \(Q\) placed at the center. So
\[ V(r)=\frac{Q}{4\pi\varepsilon_0 r} \] for \(r>R\), with relative permittivity \(1\), so \(\varepsilon_0=8.854\times10^{-12}\) F/m applies directly.

Step 4: Write the potentials at the two points.
\[ V_1=\frac{Q}{4\pi\varepsilon_0\times5},\qquad V_2=\frac{Q}{4\pi\varepsilon_0\times4} \]

Step 5: Set up the work done.
Moving a charge \(q\) from one point to another against the field takes work equal to the change in potential energy,
\[ W=q(V_2-V_1)=\frac{qQ}{4\pi\varepsilon_0}\left(\frac{1}{4}-\frac{1}{5}\right) \]

Step 6: Simplify using the special form of \(q\).
The point charge is given as
\[ q=\pi\times8.854\times10^{-12}\ \text{C} \]
so
\[ \frac{q}{4\pi\varepsilon_0}=\frac{\pi\times8.854\times10^{-12}}{4\pi\times8.854\times10^{-12}}=\frac{1}{4} \]
The \(\pi\) and \(8.854\times10^{-12}\) cancel exactly.

Step 7: Substitute the numbers.
\[ W=Q\times\frac{1}{4}\times\left(\frac{1}{4}-\frac{1}{5}\right)=32\times10^{-6}\times\frac{1}{4}\times\frac{1}{20} \]
\[ W=32\times10^{-6}\times\frac{1}{80}=0.4\times10^{-6}\ \text{J} \]

Final Answer:
\[ \boxed{0.40\ \mu\text{J}} \]