(a) An electric dipole consists of two point charges q and -q separated by a distance 2a. Derive an expression

Question

q

(b) A dipole is placed in

x

y

plane such that charges

q

and

-q

are located at

x = a

and

x = b

respectively. There exists an electric field

\vec{E} = 2\,\hat{i}\ \dfrac{\text{N}}{\text{C}}

in the region. Calculate the force

\vec{F}

and torque

\vec{\tau}

experienced by the dipole.

(OR)

(a)

E_1

and

E_2

are the emfs of two cells with internal resistances

r_1

and

r_2

respectively, connected in parallel. Deduce an expression for the equivalent emf and equivalent internal resistance of the combination.

(b) A parallel combination, as stated in

(a) above, of two cells of emfs

E

and

3E

and internal resistances

R

each is connected across a resistance

2R

. Find the current that flows through resistance

2R

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Part (a)Concept understanding — Electric Field Due to a Dipole

A short dipole's field falls off as $1/r^3$ (faster than a single charge's $1/r^2$ , since the two opposite charges' fields nearly cancel at large distances) and depends on direction: on the axial line (through the charges, extended), $E_{\text{axial}}=2kp/r^3$ , directed along $\vec{p}$ ; on the equatorial line (perpendicular bisector), $E_{\text{eq}}=kp/r^3$ , directed opposite to $\vec{p}$ -- exactly half the axial value at the same distance. At a general point making angle $\theta$ with the axis, $E=(kp …

Part (b)Concept understanding — Internal Resistance

Internal Resistance – The Battery That Fights Itself

Imagine you have a bucket of water with a tap at the bottom. When you open the tap fully, water flows out freely. But if the pipe is narrow or clogged, the flow is weaker even though the bucket is full. A battery behaves the same way: it has a "full bucket" of electrical energy (its emf, or electromotive force), but inside the battery, there is always some "clogged pipe" that resists the flow of charge. That clog is called internal resistance.

The Intuition

Every real battery is not a perfect source of voltage. If you connect a small bulb to a fresh battery, it glows brightly. But if you connect a powerful motor that draws a large current, the same battery might struggle — the voltage at its terminals drops, and the motor runs slower. Why? Because inside the battery, the chemical reactions and the materials themselves have a natural opposition to the flow of charge. This opposition is internal resistance, denoted by $r$ (or sometimes $R_{\text{int}}$ ).

Think of it this way: the battery has two "battles" to fight. First, it must push charge through the external circuit (the bulb, the motor, the wires). Second, it must push charge through its own internal structure. The harder it has to push (i.e., the larger the current), the more voltage it "loses" inside itself.

The Precise Statement

Internal resistance is the opposition to the flow of electric current offered by the materials and chemical processes inside a source of emf (like a battery, cell, or generator). It is measured in ohms ( $\Omega$ ).

For a cell or battery, the relationship between the emf ( $\mathcal{E}$ ), the terminal voltage ( $V$ ), the current ( $I$ ), and the internal resistance ( $r$ ) is given by:

V = \mathcal{E} - I r

This is the single most important equation to remember.

V = \mathcal{E} - I r

Here:

$\mathcal{E}$ is the electromotive force — the maximum voltage the battery can provide when no current flows (open circuit). Think of it as the "ideal" voltage.
$V$ is the terminal voltage — the actual voltage you measure across the battery's terminals when it is delivering current.
$I$ is the current flowing through the circuit (and therefore through the battery itself).
$r$ is the internal resistance.

The term $I r$ is the voltage drop inside the battery. It is the "price" the battery pays for delivering current.

What Happens in Different Situations?

Condition	Current $I$	Terminal Voltage $V$	Why?
Open circuit (no load)	$I = 0$	$V = \mathcal{E}$	No current means no internal drop.
Small load (e.g., a dim bulb)	Small $I$	$V \approx \mathcal{E}$	The $I r$ term is tiny.
Large load (e.g., a powerful motor)	Large $I$	$V < \mathcal{E}$ significantly	The $I r$ drop becomes noticeable.
Short circuit (wires directly across terminals)	Very large $I$	$V \approx 0$	Almost all voltage is lost inside the battery; the battery heats up and may be damaged.

Watch out

A common mistake is to think that internal resistance is a "bad" thing that can be eliminated. It cannot — every real source has some internal resistance. Even a brand new AA battery has a small $r$ (typically $0.1$ to $0.5\ \Omega$ ). Old or weak batteries have much larger $r$ , which is why they fail to power high-current devices.

Why Does Internal Resistance Matter?

Power loss inside the battery: The power dissipated as heat inside the battery is $P_{\text{loss}} = I^2 r$ . This is why batteries get warm when delivering large currents.
Maximum power transfer: There is a famous theorem (Maximum Power Transfer Theorem) that says a source delivers maximum power to an external load when the load resistance equals the internal resistance ( $R_{\text{load}} = r$ ). But this is inefficient — half the power is wasted inside the source. …

Why this formula?

Internal Resistance: Why the Key Formulas Hold

Internal resistance is a fundamental concept in real-world circuits. No battery is perfect — every practical source has some internal resistance ( $r$ ) that opposes the flow of current inside the source itself.

1. The Core Idea: A Real Battery = An Ideal Source + A Resistor

Think of a real battery as:

An ideal EMF source ( $\mathcal{E}$ ) — provides constant voltage with zero internal resistance.
A small resistor ( $r$ ) — connected in series inside the battery.

Why series? Because the current that leaves the battery must first pass through its internal material (electrolyte, electrodes), which offers resistance.

2. The Terminal Voltage Formula

When the battery delivers current $I$ to an external circuit:

The ideal source produces $\mathcal{E}$ .
The internal resistor $r$ drops some voltage: $V_{\text{drop}} = I r$ (Ohm's law).
The voltage available at the terminals ( $V$ ) is what's left:

V = \mathcal{E} - I r

Why this makes sense:

If $I = 0$ (open circuit), $V = \mathcal{E}$ — you measure the full EMF.
If $I$ increases, $V$ decreases — the internal drop grows.
If $I$ is huge (short circuit), $V \to 0$ and all voltage is lost inside.

3. The Short-Circuit Current Formula

If you connect the terminals directly (external resistance $R = 0$ ):

The only resistance in the circuit is $r$ .
By Ohm's law: $I_{\text{short}} = \frac{\mathcal{E}}{r}$

Why this holds:

The entire EMF is now dropped across $r$ alone.
This is the maximum current the battery can deliver — limited by its internal resistance.
In practice, this can damage the battery (heating, chemical damage).

4. Power Delivered to External Load

When the battery is connected to an external load $R$ :

Total circuit resistance: $R + r$
Current: $I = \frac{\mathcal{E}}{R + r}$
Power delivered to the external load ( $P_{\text{out}}$ ):

P_{\text{out}} = I^2 R = \left( \frac{\mathcal{E}}{R + r} \right)^2 R

Why this matters:

Power is not simply $\mathcal{E}^2 / R$ — because $r$ limits current. …

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Equatorial field $\vec E=-\dfrac{1}{4\pi\varepsilon_0}\dfrac{\vec p}{(r^2+a^2)^{3/2}}$ ( $\to -kp/r^3$ for $r\gg a$ ); a dipole aligned with a uniform field has $\vec F=0,\ \vec\tau=0$ .
Parallel cells: $E_{eq}=\dfrac{E_1r_2+E_2r_1}{r_1+r_2}$ , $r_{eq}=\dfrac{r_1r_2}{r_1+r_2}$ ; for the given data $E_{eq}=2E$ , $r_{eq}=R/2$ , and the current through $2R$ is $\dfrac{4E}{5R}$ .

Part (a) — Equatorial field, force and torque

Equatorial (broadside) field. Put $+q$ at $(a,0)$ , $-q$ at $(-a,0)$ ; $\vec p=2aq\,\hat i$ (from $-q$ to $+q$ ). A point $P(0,r)$ on the equatorial plane is equidistant $\sqrt{r^2+a^2}$ from both charges, so

E_+=E_-=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2+a^2}.

By symmetry the components perpendicular to the axis cancel and the axial components (each with factor $\cos\theta=\dfrac{a}{\sqrt{r^2+a^2}}$ ) add, both pointing along $-\hat i$ :

\vec E=-2\cdot\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2+a^2}\cdot\frac{a}{\sqrt{r^2+a^2}}\,\hat i=-\frac{1}{4\pi\varepsilon_0}\frac{2aq}{(r^2+a^2)^{3/2}}\,\hat i=-\frac{1}{4\pi\varepsilon_0}\frac{\vec p}{(r^2+a^2)^{3/2}}.

For $r\gg a$ , $(r^2+a^2)^{3/2}\approx r^3$ :

\vec E=-\frac{1}{4\pi\varepsilon_0}\frac{\vec p}{r^3}\quad(\text{antiparallel to }\vec p).

Force and torque. Now $+q$ at $x=a$ , $-q$ at $x=b$ , field $\vec E=2\hat i$ N C $^{-1}$ (uniform).

Net force: $\vec F=q\vec E+(-q)\vec E=0$ .
Dipole moment: $\vec p=q(a-b)\hat i$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 32 on this concept.

CBSE 2026Set A1 markMCQ
$Q. A cell of internal resistance r is connected to an external resistance R. The current will be maximum in R, if (A) R = r/2 (B) R = r (C) R > r (D) R < r$
›Reveal solutionSolution

I = ε/(R+r); smaller R ⇒ larger current, so current is maximum for R < r (ideally R → 0).

The circuit current is $I = \dfrac{\varepsilon}{R+r}$ .

For a fixed emf ε and fixed internal resistance r, the current increases as the external resistance R decreases. Hence the current through R is maximum when R is as small as possible. Among the given choices, this corresponds t …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The electromotive force of an accumulator battery is 10 V and internal resistance 0.5Ω. The maximum electric current obtained from the battery will be$
(a) 5 A
(b) 10 A
(c) 20 A
(d) 0.05 A
›Reveal solutionSolution

The maximum current a cell can deliver is its short-circuit current, I = EMF / internal resistance.

A real battery has EMF (epsilon) and internal resistance r. When connected to an external circuit of resistance R, the current is I = epsilon/(R+r), which is largest when R = 0 (sho …

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CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: The quantity measured across a cell without drawing any current from it, is ..... .$
›Reveal solutionSolution

The potential difference measured across a cell's terminals when no current is drawn from it equals the cell's EMF.

When current $I$ flows, the terminal voltage is $V = \varepsilon - Ir$ (less than the EMF $\varepsilon$ due to the voltage drop across internal resistance $r$ ). When no current is drawn ( $I=0$ , open circuit, e.g. measured with a …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The internal resistance of a cell depends on:$
(a) the area of the plates
(b) the distance between the plates
(c) the concentration of the electrolyte
(d) All of the above
›Reveal solutionSolution

A cell's internal resistance behaves like the resistance of the electrolyte column between its electrodes, so it depends on every geometric and chemical factor that affects that column.

Inside a cell, current flows through the electrolyte between the two electrodes. Treating the electrolyte as a conducting medium, its resistance follows the same rule as any conductor: $r=\rho\dfrac{l}{A}$ , where $\rho$ is the electrolyte's resistivity, $l$ the distance between the plates, and $A$ the area of the plates. So (i) a larger plate area $A$ gives a lower resistance (more parallel paths for current), (ii) a larger separation $l$ between plates gives a higher resistance (lon …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Let the point P be at distance r from the centre of the dipole on the side of the charge q, as shown in Figure, then.$
(a) E(+q) = q / [4πε₀ (r + a)²] , directed along P̂
(b) E(+q) = q / [4πε₀ (r - a)²] , directed along P̂
(c) E(+q) = -q / [4πε₀ (r + a)³] , directed along P̂
(d) None of these
›Reveal solutionSolution

The point P is closer to the $+q$ charge (distance $r-a$ ) than to $-q$ (distance $r+a$ ), so the field due to $+q$ at P follows the inverse-square law with $(r-a)$ in the denominator, pointing away from $+q$ along the dipole axis.

Figure recap: The dipole has $+q$ at one end and $-q$ at the other, separated by $2a$ , with centre $O$ . Point $P$ lies on the axial line at distance $r$ from $O$ , on the side of the $+q$ charge, so $P$ is only $(r-a)$ away from $+q$ but $(r+a)$ away from $-q$ .

Field due to $+q$ at P: Using Coulomb's law for a point charge,

$E(+q) = \frac{q}{4\pi\varepsilon_0 (r-a)^2}$

directed AWAY from $+q$ , i.e. along $\hat p$ (the unit vector along the dipole axis, pointing from $-q$ towards $+q$ and beyond, towards P).

…

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CBSE 2026Set SEM31 markMCQ
$Q. The electric field intensity due to an electric dipole at a distance r from its centre in axial position is E. If the dipole is rotated through an angle of 90° about its perpendicular axis, the magnitude of the electric field intensity at the same point will be$
(a) E
(b) E/4
(c) E/2
(d) 2E
›Reveal solutionSolution
At the same distance r, the axial field is twice the equatorial field. Rotating the dipole 90° about its perpendicular axis makes the observation point equatorial, so the field drops to E/2. Option (c).

Step 1 — recall the two standard dipole fields at distance r (r ≫ dipole size), from NCERT/CBSE Class 12 Physics:

Axial (end-on): E_axial = (1/4πε₀)(2p/r³)

Equatorial (broadside): E_equatorial = (1/4πε₀)(p/r³) …
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CBSE 2025Set ANNUAL1 mark
$Q. The electric field intensity at axis due to an electric dipole is inversely proportional to the ____________ of distance.$
›Reveal solutionSolution

The axial field of a dipole falls off much faster than that of a single point charge — as the inverse cube, not the inverse square, of distance.

For a short electric dipole of moment $p = q(2a)$ , the electric field at a point on the axial line at distance $r$ from the centre (for $r \gg a$ ) is:

$E_{axial} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The maximum current that can be drawn from a cell is when$
(a) R = 0
(b) r = 0
(c) R > r
(d) r > R
›Reveal solutionSolution

Current from a cell $I = E/(R+r)$ is largest when the total resistance $(R+r)$ is smallest, i.e. when the external resistance R = 0.

For a cell of emf E and internal resistance r connected to an external resistance R, the current is

$I = \frac{E}{R+r}$

…

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CBSE 2024Set 55/5/11 markMCQ
$2\,\Omega$
›Reveal solutionSolution
A real battery has an internal resistance that causes its terminal voltage to drop when current is drawn. By analyzing the battery's behavior under two different load conditions, we can determine its internal resistance, which is $\boxed{0.5\,\Omega}$ .

A battery is not an ideal voltage source; it possesses an inherent internal resistance, denoted by $r$ . This internal resistance is effectively in series with the battery's electromotive force (EMF), $E$ . When a current $I$ is drawn from the battery through an external resistor $R$ , a voltage drop occurs across this internal resistance, equal to $Ir$ .

The voltage available across the terminals of the battery, known as the terminal voltage $V$ , is therefore less than the EMF $E$ . It is given by:
$V = E - Ir$
This terminal voltage is also the voltage across the external resistor $R$ , so $V = IR$ .

Equating these two expressions for $V$ , we get:
$IR = E - Ir$
Rearranging this equation to solve for the EMF $E$ :
$E = I(R + r)$

Formulate equations for each scenario.

We are given two distinct situations where the battery is connected to a different external resistor, resulting in a different current. We will apply the formula $E = I(R + r)$ to each case.

Scenario 1: The battery supplies a current $I_1 = 0.9\,\text{A}$ through an external resistor $R_1 = 2\,\Omega$ . Using the formula $E = I(R + r)$ :

$E = 0.9\,\text{A} \times (2\,\Omega + r)$ $E = 1.8 + 0.9r \quad \text{(Equation 1)}$
* **Scenario 2:** The battery supplies a current $I_2 = 0.3\,\text{A}$ through an external resistor $R_2 = 7\,\Omega$. Using the formula $E = I(R + r)$:
$E = 0.3\,\text{A} \times (7\,\Omega + r)$ $E = 2.1 + 0.3r \quad \text{(Equation 2)}$

Solve the system of equations for $r$ . …
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CBSE 2024Set 55/1/11 markMCQ
$6\ \text{V}$
›Reveal solutionSolution
Two cells with internal resistances in parallel drive current through each other; the terminal voltage $V_{AB}$ is found by treating each branch as a source with EMF and internal resistance, then using the parallel-source formula. The answer is $\boxed{8 \text{ V}}$ .

Why this approach works: cells with internal resistance in parallel

When two cells are connected in parallel between the same two points A and B, they don't simply "add up" like batteries in series. Each cell has an EMF (electromotive force) and an internal resistance, and they can drive current through each other. The stronger cell (higher EMF per unit resistance) will actually charge the weaker one.

The key insight: treat each branch as a source characterized by its EMF $\mathcal{E}$ and internal resistance $r$ . The terminal voltage $V_{AB}$ across the parallel combination is the voltage that appears at the terminals when both sources are connected together. This is given by the weighted average of the EMFs, where the weights are the conductances (reciprocals of internal resistances).
$V_{AB} = \frac{\frac{\mathcal{E}_1}{r_1} + \frac{\mathcal{E}_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}$
This formula comes from applying Kirchhoff's laws: the current from each source adjusts so that both branches have the same terminal voltage.

Step-by-step solution

1. Identify the parameters of each branch from the circuit

Figure: two-branch circuit between A and B

Reading the circuit:

Upper branch: EMF $\mathcal{E}_1 = 12 \text{ V}$ , internal resistance $r_1 = 1 \, \Omega$

Lower branch: EMF $\mathcal{E}_2 = 6 \text{ V}$ , internal resistance $r_2 = 0.5 \, \Omega$

2. Calculate the "weighted EMFs" (EMF divided by internal resistance)

These represent the short-circuit current each source can deliver:
$\frac{\mathcal{E}_1}{r_1} = \frac{12}{1} = 12 \text{ A}$ $\frac{\mathcal{E}_2}{r_2} = \frac{6}{0.5} = 12 \text{ A}$
Interestingly, both branches have the same short-circuit current capability.

3. Calculate the sum of conductances …
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CBSE 2023Set 55/1/11 markMCQ
$r$
(a) $\dfrac{F}{16}$
(b) $\dfrac{F}{8}$
(c) $\dfrac{F}{4}$
(d) $\dfrac{F}{2}$
›Reveal solutionSolution
The force on a point charge due to a short electric dipole on its axis follows an inverse-cube law. Doubling the distance reduces the force by a factor of 8, so the new force is $F/8$ .

The key here is understanding how the electric field of a dipole behaves with distance. A short electric dipole (two equal and opposite charges separated by a small distance) does not produce a field that falls off like a point charge ( $1/r^2$ ). Instead, along its axis, the field falls off as $1/r^3$ . This is because the fields from the two opposite charges nearly cancel at large distances, leaving a weaker, faster-decaying net field.

Since force on a test charge is $F = qE$ , and the test charge itself doesn’t change, the force is directly proportional to the dipole’s electric field at that point. So if the field changes by a factor, the force changes by the same factor.

Let’s work through it step by step.

Recall the formula for the axial field of a short dipole. For a dipole of dipole moment $p$ , at a point on its axis at distance $r$ from its centre (where $r$ is much larger than the separation between the two charges), the electric field magnitude is:

$E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2p}{r^3}$
This is a standard result — the $1/r^3$ dependence is the hallmark of a dipole field.

Relate force to field. The force on a point charge $q$ placed in this field is simply:

$F = qE = q \cdot \frac{1}{4\pi\varepsilon_0} \cdot \frac{2p}{r^3}$
So $F \propto \frac{1}{r^3}$ .

Now double the distance. Let the initial distance be $r$ , giving force $F$ . …
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CBSE 2023Set 55/1/11 markMCQ
$8\ \text{V}$
(a) $4\ \Omega$
(b) $3\ \Omega$
(c) $2\ \Omega$
(d) $1\ \Omega$
›Reveal solutionSolution
The key idea is that the open-circuit voltage is the cell’s EMF ( $\mathcal{E} = 8\ \text{V}$ ), and the drop to $4\ \text{V}$ when $4\ \text{A}$ flows is due entirely to the voltage drop across the internal resistance $r$ . Using $V = \mathcal{E} - Ir$ , we get $r = 1\ \Omega$ , so the correct option is (d).

Every real cell behaves like a perfect EMF source $\mathcal{E}$ in series with a small internal resistance $r$ . When no current flows (open circuit), the terminal voltage equals $\mathcal{E}$ — there’s no drop across $r$ . But the moment you draw current, $r$ steals some voltage: $V_{\text{terminal}} = \mathcal{E} - I r$ . That’s the whole physics in one line.

Here, the open-circuit reading gives $\mathcal{E} = 8\ \text{V}$ . When $4\ \text{A}$ is drawn, the terminal voltage crashes to $4\ \text{V}$ . That $4\ \text{V}$ loss is $I r$ . So:

Write the terminal voltage equation:

$V = \mathcal{E} - I r$

Plug in the numbers:

$4 = 8 - (4) r$

Solve for $r$ : $4r = 8 - 4 = 4 \quad\Rightarrow\quad r = 1\ \Omega$ …
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Part (a) — Equatorial field, force and torque

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