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 via superposition, vector form)

Step 1: An electric dipole is two charges \(+q\) and \(-q\) a distance \(2a\) apart. The dipole moment is \(\vec{p}=2qa\), a vector pointing from \(-q\) to \(+q\), measured in C m. It measures how strong and how 'stretched' the charge pair is.
Step 2: Take the dipole axis as the x-axis with centre at O. A point P lies on this axis at distance \(r\). Both charge fields lie along the axis, so we may add magnitudes with sign.
Step 3: The near charge (\(+q\), distance \(r-a\)) pushes outward with \(E_+ = \dfrac{kq}{(r-a)^2}\) and the far charge (\(-q\), distance \(r+a\)) pulls inward with \(E_- = \dfrac{kq}{(r+a)^2}\), where \(k = \dfrac{1}{4\pi\varepsilon_0}\).
Step 4: Net field \(E = E_+ - E_- = kq\left[\dfrac{(r+a)^2-(r-a)^2}{(r^2-a^2)^2}\right] = kq\cdot\dfrac{4ar}{(r^2-a^2)^2}\). With \(p=2qa\), \(\;E = \dfrac{k\,2pr}{(r^2-a^2)^2}\). \[ \boxed{\,E_{axial} = \frac{1}{4\pi\varepsilon_0}\frac{2pr}{(r^2-a^2)^2}\;\xrightarrow{r\gg a}\;\frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}\,} \] pointing along \(\vec{p}\). Note it is twice the equatorial field and falls as \(1/r^3\).

OPTION 2 (Same network solved by charge and voltage sharing)

Step 1: Compare the potential drops. The mid-point of the 4-2 arm and the mid-point of the 2-1 arm sit at the same potential because \(4/2 = 2/1\). Equal potential across the 12 \(\mu\)F means \(Q_{12} = C\,\Delta V = 12\times 0 = 0\); it is dead and can be deleted.
Step 2: Series pair 4 and 2: reciprocal add \(\frac{1}{C_I} = \frac{1}{4}+\frac{1}{2} = \frac{3}{4}\), so \(C_I = \frac{4}{3}\,\mu\)F. Series pair 2 and 1: \(\frac{1}{C_{II}} = \frac{1}{2}+1 = \frac{3}{2}\), so \(C_{II} = \frac{2}{3}\,\mu\)F.
Step 3 (i): Parallel branches add directly: \(C = \frac{4}{3}+\frac{2}{3} = 2\,\mu\)F. \[ \boxed{C = 2\ \mu\text{F}} \]
Step 4 (ii): Voltage across the 4-2 branch is 22 V. Its series charge \(q = C_I V = \frac{4}{3}\times 22 = 29.3\,\mu\)C is common to both 4 \(\mu\)F and 2 \(\mu\)F. \[ \boxed{q_{4\mu F} \approx 29.3\ \mu\text{C}} \]
Step 5 (iii): Using total charge \(Q = CV = 2\times 22 = 44\,\mu\)C, the energy is \(U = \tfrac{1}{2}QV = \tfrac{1}{2}(44\times10^{-6})(22) = 4.84\times10^{-4}\) J. \[ \boxed{U = 484\ \mu\text{J}} \] which matches \(\tfrac12 CV^2\).