Q.A conducting sphere of radius $10\,\text{cm}$ has an unknown charge. If the electric field $20\,\text{cm}$ from the centre of the sphere is $1.5 \times 10^{3}\,\text{N/C}$ and points radially inward, what is the net charge on the sphere?

Concept: Gauss’s Law — the net electric flux through a closed surface equals $\frac{Q_{\text{enc}}}{\varepsilon_0}$.

Step 1: Choose a spherical Gaussian surface of radius $r = 20\ \text{cm} = 0.2\ \text{m}$, concentric with the conducting sphere. The field is radial and uniform over this surface.

Step 2: The electric flux is

Given $E = 1.5 \times 10^{3}\ \text{N/C}$ (inward, so negative if outward is positive), but magnitude is enough — sign will come from direction.

Step 3: By Gauss’s law,

Substitute values:

$Q = (8.85 \times 10^{-12}) \times (1.5 \times 10^{3}) \times 4\pi \times (0.2)^2$ …

The key idea is that for a conducting sphere, any excess charge resides entirely on its surface. Outside the sphere, the electric field behaves exactly as if all that charge were concentrated at the centre. This is a direct consequence of spherical symmetry and Gauss’s law.

Why does this matter? Because it means we can treat the sphere as a point charge when calculating the field at any point outside it. The problem gives us the field at a distance of $20\,\text{cm}$ from the centre — that’s outside the sphere (radius $10\,\text{cm}$), so the point-charge model is valid.

The field points radially inward. That’s a crucial detail: it tells us the charge is negative. A positive charge would produce an outward field.

Now let’s work through the calculation.

1. Identify the relevant distance.

   The sphere’s radius is $R = 10\,\text{cm} = 0.10\,\text{m}$.

   The point where the field is given is $r = 20\,\text{cm} = 0.20\,\text{m}$ from the centre.

   Since $r > R$, we are outside the sphere.

2. Apply Gauss’s law for a spherical Gaussian surface.

   For a spherically symmetric charge distribution, the electric field at distance $r$ from the centre is:

where $q$ is the net charge enclosed. For a conducting sphere, all charge is on the surface, so the enclosed charge is just the net charge on the sphere.

3. Plug in the known values. We have $E = 1.5 \times 10^{3}\,\text{N/C}$ and $r = 0.20\,\text{m}$. The constant $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}\,\text{N·m}^2/\text{C}^2$. So:

4. Solve for $|q|$. First, $(0.20)^2 = 0.04$. Then: $|q| = \frac{1.5 \times 10^{3} \times 0.04}{9 \times 10^{9}}$ …

Method: Gauss's Law

Why Gauss's Law?

For a spherically symmetric charge distribution, the electric field at any point depends only on the net charge enclosed inside a spherical Gaussian surface. This lets us find the charge without knowing the internal distribution.

Steps

Step 1: Choose a Gaussian surface

Draw a spherical surface of radius $r = 20\ \text{cm}$ (where the field is given), concentric with the conducting sphere.

Step 2: Apply Gauss's Law

Gauss's Law states:

For a spherical surface, $\vec{E}$ is radial and constant in magnitude over the surface, so:

Step 3: Identify the enclosed charge

The Gaussian surface at $20\ \text{cm}$ encloses the entire conducting sphere (radius $10\ \text{cm}$). So $Q_{\text{enc}}$ is the net charge on the sphere — this is what we need.

Step 4: Solve for $Q$

Step 5: Plug in values

• $r = 20\ \text{cm} = 0.20\ \text{m}$
• $E = 1.5 \times 10^{3}\ \text{N/C}$ (inward direction)
• $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/\text{N·m}^2$ …

1. ✗ Mistake: Forgetting the sign of the charge

What students do wrong:

They compute the magnitude of charge correctly but write the answer as positive.

Why it’s wrong:

The electric field points radially inward. For a spherical Gaussian surface, an inward field means the net flux is negative (field lines enter the surface). By Gauss’s law:

If $\Phi_E < 0$, then $q_{\text{enc}} < 0$.

✓ How to avoid:

Always check the direction of the field relative to the outward normal of your Gaussian surface.

• Outward field → positive charge
• Inward field → negative charge

Here, inward field → negative charge.

2. ✗ Mistake: Using the wrong radius for the Gaussian surface

What students do wrong:

They take the Gaussian surface radius as $10\,\text{cm}$ (the sphere’s radius) instead of $20\,\text{cm}$ (where the field is given).

Why it’s wrong:

Gauss’s law requires you to draw a closed surface that passes through the point where you know the field. The field is given at $20\,\text{cm}$, so your Gaussian sphere must have radius $r = 20\,\text{cm} = 0.2\,\text{m}$.

✓ How to avoid:

• The Gaussian surface radius = distance from centre to the point where $E$ is known.
• The sphere’s own radius only matters to decide if the point is inside or outside the conductor.

Here, $20\,\text{cm} > 10\,\text{cm}$, so the point is outside — the sphere behaves like a point charge at the centre.

3. ✗ Mistake: Using the formula for a point charge incorrectly

What students do wrong:

They write $E = \frac{kq}{r^2}$ but forget to convert cm to metres, or they use $r = 10\,\text{cm}$.

✓ How to avoid:

Always convert to SI units:

• $r = 20\,\text{cm} = 0.20\,\text{m}$
• Use $k = 9 \times 10^9\ \text{N m}^2/\text{C}^2$ or $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/\text{N m}^2$

Then:

Plug in:

So $q = -6.67 \times 10^{-9}\ \text{C}$ (negative because field is inward).

4. ✗ Mistake: Forgetting that the sphere is a conductor

What students do wrong:

They worry about charge distribution inside the sphere.

Why it’s wrong: …