Q.A uniformly charged conducting sphere of $2.4\,\text{m}$ diameter has a surface charge density of $80.0\,\mu\text{C/m}^2$.

Here's how to approach this problem, focusing on the underlying concepts for each part.

(a) Find the charge on the sphere.

The key idea here is the definition of surface charge density.

1. A uniformly charged conducting sphere holds its charge entirely on its surface.
2. Surface charge density ($\sigma$) is defined as the total charge ($Q$) divided by the surface area ($A$) over which it is distributed: $\sigma = Q/A$.
3. The surface area of a sphere with diameter $D$ is $A = \pi D^2$.
4. We are given $D = 2.4\,\text{m}$ and $\sigma = 80.0\,\mu\text{C/m}^2 = 80.0 \times 10^{-6}\,\text{C/m}^2$.
5. Therefore, $Q = \sigma A = \sigma (\pi D^2)$.

$Q = (80.0 \times 10^{-6}) \times \pi (5.76)\,\text{C}$ …

When dealing with charged objects, two fundamental quantities are often sought: the total charge and the electric flux. For a uniformly charged sphere, these calculations become straightforward due to its high degree of symmetry.

Concept and Intuition

1. Surface Charge Density ($\sigma$): This quantity tells us how much charge is packed onto each unit area of a surface. If a charge $Q$ is spread uniformly over a surface area $A$, then the surface charge density is simply $Q/A$. Conversely, if we know $\sigma$ and $A$, we can find the total charge $Q = \sigma A$. For a sphere, the surface area is $4\pi R^2$.

2. Gauss's Law: This is one of Maxwell's equations and a cornerstone of electrostatics. It provides a powerful way to calculate electric flux, especially for symmetric charge distributions. Gauss's Law states that the total electric flux ($\Phi_E$) through any closed surface (called a Gaussian surface) is directly proportional to the total electric charge ($Q_{enc}$) enclosed within that surface.

   $\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$

   Here, $\epsilon_0$ is the permittivity of free space, a fundamental constant.

   For a uniformly charged sphere, the electric field lines point radially outward (if positive charge) or inward (if negative charge). If we choose a spherical Gaussian surface concentric with the charged sphere, the electric field $\vec{E}$ will be uniform in magnitude and perpendicular to the surface everywhere. This simplifies the flux integral significantly, making Gauss's Law the ideal tool for this problem.

Let's solve the problem step-by-step.

Given Data:

• Diameter of the sphere, $D = 2.4\,\text{m}$
• Surface charge density, $\sigma = 80.0\,\mu\text{C/m}^2$

We need to convert the surface charge density to standard SI units:

$\sigma = 80.0 \times 10^{-6}\,\text{C/m}^2$

(a) Find the charge on the sphere.

1. Calculate the radius of the sphere:

   The radius $R$ is half of the diameter.

   $R = \frac{D}{2} = \frac{2.4\,\text{m}}{2} = 1.2\,\text{m}$

2. Calculate the surface area of the sphere:

   The surface area $A$ of a sphere is given by the formula $A = 4\pi R^2$.

   $A = 4\pi (1.2\,\text{m})^2$

   $A = 4\pi (1.44\,\text{m}^2)$

   $A = 5.76\pi\,\text{m}^2$

3. Calculate the total charge on the sphere:

   The total charge $Q$ is the product of the surface charge density $\sigma$ and the surface area $A$.

   $Q = \sigma A$

   $Q = (80.0 \times 10^{-6}\,\text{C/m}^2) \times (5.76\pi\,\text{m}^2)$

   $Q = 460.8\pi \times 10^{-6}\,\text{C}$

   Now, substitute the numerical value of $\pi \approx 3.14159$:

   $Q \approx 460.8 \times 3.14159 \times 10^{-6}\,\text{C}$

   $Q \approx 1447.645 \times 10^{-6}\,\text{C}$

   $Q \approx 1.447645 \times 10^{-3}\,\text{C}$

   Rounding to three significant figures (consistent with $80.0\,\mu\text{C/m}^2$):

   $Q \approx 1.45 \times 10^{-3}\,\text{C}$

--- …

Method: Gauss’s Law

Why this method?

Gauss’s Law directly relates the total electric flux through a closed surface to the enclosed charge. For a conducting sphere, all charge resides on the surface, making the spherical Gaussian surface coincide with the sphere itself.

Step 1 — Find the charge on the sphere

Given:

• Diameter = $2.4\,\text{m}$ → radius $r = 1.2\,\text{m}$
• Surface charge density $\sigma = 80.0\,\mu\text{C/m}^2 = 80.0 \times 10^{-6}\,\text{C/m}^2$

Charge on a surface is:

Surface area of sphere = $4\pi r^2$

Step 2 — Find total electric flux leaving the surface

Gauss’s Law:

$\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$ …

Common Mistakes & How to Avoid Them — Gauss Law (Conducting Sphere)

Mistake 1: Confusing radius with diameter

The error:

Students often take $2.4\,\text{m}$ as the radius and plug it directly into $A = 4\pi r^2$.

Why it happens:

The problem says "diameter" but many read it as "radius" out of habit.

How to avoid:

• Always underline the word "diameter" or "radius" in the question.
• Write:

• Then use $r = 1.2\,\text{m}$ in all formulas.

Mistake 2: Forgetting to convert $\mu\text{C}$ to $\text{C}$

The error:

Using $\sigma = 80.0$ directly without converting to coulombs.

Why it happens:

$\mu\text{C}$ is common in problems, but Gauss Law formulas require SI units.

How to avoid:

• Always convert microcoulombs:

• Write the conversion step explicitly before any calculation.

Mistake 3: Using the wrong formula for total charge

The error:

Using $Q = \sigma \cdot (4\pi r^2)$ but with wrong $r$, or using $Q = \sigma \cdot (2\pi r^2)$.

Why it happens:

Confusing surface area of a sphere ($4\pi r^2$) with other shapes.

How to avoid:

• Memorise: Surface area of sphere = $4\pi r^2$ (not $2\pi r^2$, not $\pi r^2$).
• Write the formula clearly:

Mistake 4: Forgetting that flux = $\frac{Q}{\varepsilon_0}$ for a closed surface

The error:

Trying to compute flux by integrating $E \cdot dA$ manually, or using $E = \frac{\sigma}{\varepsilon_0}$ incorrectly.

Why it happens:

Students forget Gauss's Law:

How to avoid:

• Always recall: For any closed surface, total flux depends only on enclosed charge.
• For a conducting sphere, all charge is on the surface, so $Q_{\text{enc}} = Q$ (from part a).
• Then:

No need to compute $E$ or integrate.

Mistake 5: Using $\varepsilon_0$ incorrectly or with wrong units

The error:

Using $\varepsilon_0 = 8.85 \times 10^{-12}$ but forgetting units, or using $8.85 \times 10^{-12}\,\text{F/m}$ without realising it's $\text{C}^2/\text{N·m}^2$.

How to avoid:

• Write units explicitly in every step: …