Two metal spheres, one of radius $R$ and the other of radius $2R$, both have same surface charge density $\sigma$. They are brought in contact and separated. What will be new surface charge densities on them?

Two metal spheres, one of radius R and the other of radius 2R, both have same surface charge density σ. They a

NCERT Exemplar · Q29

R

Uttarakhand UbseLong· 3mImportance★★★★★

89% · 50/56 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

Contact equalises the potentials ( $q\propto r$ ) and conserves total charge, giving $q_1=\tfrac{20}{3}\pi\sigma R^2$ , $q_2=\tfrac{40}{3}\pi\sigma R^2$ , hence $\sigma_1'=\tfrac{5}{3}\sigma$ (small sphere) and $\sigma_2'=\tfrac{5}{6}\sigma$ (large sphere).

Surface charge density alone does not fix the final state — potential does. When two conductors touch, charge flows until their potentials are equal, and for a sphere $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ depends on both charge and radius.

1. Initial charges

Q_1=\sigma\cdot4\pi R^2=4\pi\sigma R^2,\qquad Q_2=\sigma\cdot4\pi(2R)^2=16\pi\sigma R^2.

Q_{\text{total}}=Q_1+Q_2=20\pi\sigma R^2.

2. Equal potential after contact

\frac{q_1}{R}=\frac{q_2}{2R}\implies q_2=2q_1.

The larger sphere ends up with twice — not four times — the charge, because $V\propto q/r$ .

3. Charge conservation

q_1+q_2=20\pi\sigma R^2\implies q_1+2q_1=20\pi\sigma R^2\implies q_1=\frac{20}{3}\pi\sigma R^2,

q_2=2q_1=\frac{40}{3}\pi\sigma R^2.

4. New surface charge densities …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ28 — (a) In a quark model of elementary particles, a neutron is made of one…→Next questionQ30 — In a circuit a 9 V cell is connected in the top line first to a capaci…

Q.Two metal spheres, one of radius $R$ and the other of radius $2R$ , both have same surface charge density $\sigma$ . They are brought in contact and separated. What will be new surface charge densities on them?

Concept understanding — Electric Potential

Electric Potential

The Idea

Potential Due to a Point Charge

Potential Difference

Relation Between Field and Potential

Superposition

Equipotential Surfaces

1. Initial charges

2. Equal potential after contact

3. Charge conservation

4. New surface charge densities …

Method: Charge Redistribution Between Conductors Brought Into Contact

Steps

Q.Two metal spheres, one of radius RRR and the other of radius 2R2R2R, both have same surface charge density σ\sigmaσ. They are brought in contact and separated. What will be new surface charge densities on them?

1. Initial charges

2. Equal potential after contact

3. Charge conservation

4. New surface charge densities …

🎓Unlock everything free for 14 days

Q.Two metal spheres, one of radius $R$ and the other of radius $2R$ , both have same surface charge density $\sigma$ . They are brought in contact and separated. What will be new surface charge densities on them?

Unlock everything free for 14 days