Question

Option 1: Write down the definition of electric dipole moment. Derive the formula for the electric field intensity at an axial point of an electric dipole.
OR
Option 2: In the given combination of capacitors (connected across a 22 V supply), find out (i) the total capacitance, (ii) the charge on the 4 \( \mu\text{F} \) capacitor and (iii) the total stored energy. The network contains capacitors of 4 \( \mu\text{F} \), 2 \( \mu\text{F} \), 1 \( \mu\text{F} \), 2 \( \mu\text{F} \), 1 \( \mu\text{F} \) and 12 \( \mu\text{F} \).

Hint:

For the dipole, add the fields of \(+q\) and \(-q\) along the axis and use \(p=2qa\) to get \(E=\frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}\). For the network, note \(4/2=2/1\) so the bridge is balanced and the 12 \(\mu\)F carries no charge; then \((4\ \text{ser}\ 2)\parallel(2\ \text{ser}\ 1)=2\ \mu\)F.

Answer 1

OPTION 1 (Axial field of an electric dipole)

Step 1: Definition of electric dipole moment. An electric dipole is a pair of equal and opposite charges \( +q \) and \( -q \) separated by a small distance \( 2a \). Its electric dipole moment is \[ \vec{p} = q\,(2a) \] a vector of magnitude \( 2qa \) directed from the negative charge to the positive charge. SI unit: coulomb-metre (C m).
Step 2: Geometry of the axial point. Let O be the centre of the dipole and P a point on the axis at distance \( r \) from O. Then \( +q \) is at distance \( (r-a) \) from P and \( -q \) at distance \( (r+a) \).
Step 3: Field due to each charge. Using \( E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{(\text{distance})^{2}} \):
Field due to \( +q \) (directed from \( +q \) towards P, i.e. away from dipole): \[ E_{+} = \frac{1}{4\pi\varepsilon_0}\frac{q}{(r-a)^{2}} \]
Field due to \( -q \) (directed from P towards \( -q \), i.e. towards dipole): \[ E_{-} = \frac{1}{4\pi\varepsilon_0}\frac{q}{(r+a)^{2}} \]
Step 4: Net field. Along the axis these are opposite, and \( E_+ > E_- \), so the resultant points along \( \vec{p} \): \[ E_{axial} = E_{+} - E_{-} = \frac{q}{4\pi\varepsilon_0}\left[\frac{1}{(r-a)^{2}} - \frac{1}{(r+a)^{2}}\right] \]
Step 5: Simplify. \[ E_{axial} = \frac{q}{4\pi\varepsilon_0}\cdot\frac{(r+a)^{2}-(r-a)^{2}}{(r^{2}-a^{2})^{2}} = \frac{q}{4\pi\varepsilon_0}\cdot\frac{4ar}{(r^{2}-a^{2})^{2}} \] Writing \( 2qa = p \): \[ \boxed{\,E_{axial} = \frac{1}{4\pi\varepsilon_0}\cdot\frac{2pr}{(r^{2}-a^{2})^{2}}\,} \]
Step 6: Short (point) dipole. For \( r \gg a \), \( (r^{2}-a^{2})\approx r^{2} \): \[ \boxed{\,E_{axial} = \frac{1}{4\pi\varepsilon_0}\cdot\frac{2p}{r^{3}}\,} \] directed along the dipole moment \( \vec{p} \).

OPTION 2 (Balanced-bridge capacitor network, 22 V)

Step 1: Recognise a Wheatstone bridge of capacitors. The six capacitors form a bridge: one arm has 4 \( \mu\text{F} \) then 2 \( \mu\text{F} \); the parallel arm has 2 \( \mu\text{F} \) then 1 \( \mu\text{F} \); and the 12 \( \mu\text{F} \) (together with the remaining 1 \( \mu\text{F} \)) bridges the two mid-points.
Step 2: Test the balance condition. For a capacitor bridge, the bridge arm carries no charge when the ratios of the arm capacitances match: \[ \frac{C_1}{C_2} = \frac{C_3}{C_4}\;\Rightarrow\; \frac{4}{2} = \frac{2}{1} = 2 \] The condition is satisfied, so the bridge is balanced and the 12 \( \mu\text{F} \) capacitor carries no charge. It may be removed.
Step 3: Reduce the two branches.
Branch 1 ( 4 \( \mu\text{F} \) in series with 2 \( \mu\text{F} \) ): \[ C_{I} = \frac{4\times 2}{4+2} = \frac{8}{6} = \frac{4}{3}\ \mu\text{F} \]
Branch 2 ( 2 \( \mu\text{F} \) in series with 1 \( \mu\text{F} \) ): \[ C_{II} = \frac{2\times 1}{2+1} = \frac{2}{3}\ \mu\text{F} \]
Step 4 (i): Total capacitance. The two branches are in parallel: \[ C = C_{I} + C_{II} = \frac{4}{3} + \frac{2}{3} = \frac{6}{3} = \boxed{2\ \mu\text{F}} \]
Step 5 (ii): Charge on the 4 \( \mu\text{F} \) capacitor. Branch 1 ( \( 4/3\ \mu\text{F} \) ) is connected across the full 22 V, so the charge stored in that series branch is \[ q_{I} = C_{I}\,V = \frac{4}{3}\times 22 = \frac{88}{3} \approx 29.3\ \mu\text{C} \] In a series branch every capacitor carries the same charge, hence \[ \boxed{\,q_{4\mu F} = \frac{88}{3}\ \mu\text{C} \approx 29.3\ \mu\text{C}\,} \] ( the voltage across the 4 \( \mu\text{F} \) is then \( 29.3/4 \approx 7.3 \) V ).
Step 6 (iii): Total stored energy. \[ U = \frac{1}{2}CV^{2} = \frac{1}{2}\times(2\times10^{-6})\times(22)^{2} \] \[ U = \frac{1}{2}\times 2\times10^{-6}\times 484 = 484\times10^{-6}\ \text{J} \] \[ \boxed{\,U = 4.84\times10^{-4}\ \text{J} = 484\ \mu\text{J}\,} \]