Consider two conducting spheres of radii $R_1$ and $R_2$ with $R_1 > R_2$. If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one.

Consider two conducting spheres of radii R₁ and R₂ – Class 12 Physics

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Concept understanding — Electric Potential

Electric Potential

The Idea

When you lift a book onto a shelf you do work against gravity, and the book gains gravitational potential energy that depends on where it sits. Electric charges behave the same way in an electric field. Electric potential is the electrical analogue of "height" in a gravitational field: it tells you the potential energy a unit charge would have at a given point.

Formally, the electric potential at a point is the work done in bringing a unit positive charge from infinity to that point, slowly (without giving it kinetic energy), against the electric field.

V = \frac{W}{q_0}

Here $W$ is the work done to move a small test charge $q_0$ from infinity to the point. Because both $W$ and $q_0$ are scalars, electric potential is a scalar quantity — it has magnitude but no direction. Its SI unit is the volt:

1\ \text{V} = 1\ \text{J/C}

Potential Due to a Point Charge

For a single point charge $q$ , the potential at a distance $r$ from it is

V = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{r}

Notice it falls off as $1/r$ , whereas the electric field of a point charge falls off as $1/r^2$ . The potential is taken as zero at infinity, our chosen reference. A positive charge makes the potential around it positive; a negative charge makes it negative.

Potential Difference

Usually we care about the potential difference between two points $A$ and $B$ :

V_A - V_B = \frac{W_{B\to A}}{q_0}

the work per unit charge needed to move a charge from $B$ to $A$ . This is the quantity a voltmeter reads and the "voltage" that drives current in a circuit.

Relation Between Field and Potential

Field and potential are two views of the same thing. The electric field is the negative rate of change of potential with distance:

E = -\frac{dV}{dr}

The minus sign says the field points from high potential toward low potential — a positive charge, left free, rolls "downhill" in potential. Where the potential changes steeply, the field is strong.

Superposition

Because potential is a scalar, the potential due to several charges is just the algebraic sum of their individual potentials — no vectors, no angles:

V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_1} + \frac{q_2}{r_2} + \cdots\right)

This makes potential far easier to compute than the field, which needs vector addition. Once you have $V$ everywhere, you can get the field by differentiating.

Equipotential Surfaces

An equipotential surface is a surface on which the potential has the same value everywhere.

No work is done in moving a charge along an equipotential surface (since $\Delta V = 0$ ).
The electric field is always perpendicular to an equipotential surface. …

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For two spheres at the same potential, the surface charge density is inversely proportional to the radius — so the smaller sphere has a higher charge density than the larger one.

The key here is to connect potential to charge density without getting lost in algebra. Let’s build the intuition first.

When two conductors are at the same potential, a charge placed on either sphere would have the same electrical potential energy per unit charge. For an isolated conducting sphere, the potential at its surface (taking infinity as zero) is simply $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$ . This comes from treating the sphere as a point charge at its centre — a standard result from Gauss’s law.

Now, charge density $\sigma$ is charge per unit area: $\sigma = \frac{Q}{4\pi R^2}$ . So the question becomes: if $V$ is fixed, how does $\sigma$ depend on $R$ ?

Let’s work through it step by step.

Write the potential condition. Both spheres are at the same potential $V$ . So:

V = \frac{1}{4\pi\epsilon_0} \frac{Q_1}{R_1} = \frac{1}{4\pi\epsilon_0} \frac{Q_2}{R_2}

The constant $\frac{1}{4\pi\epsilon_0}$ cancels, giving:

\frac{Q_1}{R_1} = \frac{Q_2}{R_2} \quad \Rightarrow \quad Q_1 = \frac{R_1}{R_2} Q_2

Since $R_1 > R_2$ , we have $Q_1 > Q_2$ — the larger sphere indeed holds more charge at the same potential, as the problem states.

Express charge density. Surface charge density is:

\sigma_1 = \frac{Q_1}{4\pi R_1^2}, \quad \sigma_2 = \frac{Q_2}{4\pi R_2^2}

Find the ratio of charge densities. Using $Q_1 = \frac{R_1}{R_2} Q_2$ :

\frac{\sigma_1}{\sigma_2} = \frac{Q_1 / R_1^2}{Q_2 / R_2^2} = \frac{(R_1 Q_2 / R_2) / R_1^2}{Q_2 / R_2^2} = \frac{Q_2 / (R_1 R_2)}{Q_2 / R_2^2} = \frac{R_2}{R_1}

So:

\sigma_1 : \sigma_2 = R_2 : R_1

Interpret the result. …

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Q.Consider two conducting spheres of radii $R_1$ and $R_2$ with $R_1 > R_2$ . If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one.

Concept understanding — Electric Potential

Electric Potential

The Idea

Potential Due to a Point Charge

Potential Difference

Relation Between Field and Potential

Superposition

Equipotential Surfaces

Method: Linking Charge Density to Potential for an Isolated Conducting Sphere

Steps

Q.Consider two conducting spheres of radii R1R_1R1​ and R2R_2R2​ with R1>R2R_1 > R_2R1​>R2​. If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one.

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Q.Consider two conducting spheres of radii $R_1$ and $R_2$ with $R_1 > R_2$ . If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one.

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