Q.Two charges $\pm 10\,\mu\text{C}$ are placed $5.0\,\text{mm}$ apart. Determine the electric field at

For a short dipole ($2a=5.0\,\text{mm}\ll r=15\,\text{cm}$) use the far‑field formulas. The dipole moment is

with $k=9\times10^{9}$ and $r^3=(0.15)^3=3.375\times10^{-3}\,\text{m}^3$.

1. Axial point P.

   $E_P=\frac{2kp}{r^3}=\frac{2(9\times10^{9})(5.0\times10^{-8})}{3.375\times10^{-3}}=2.7\times10^{5}\,\text{N/C},$

   directed along the dipole moment (from $-q$ to $+q$).
2. Equatorial point Q. …

Setup. The charges are $q=10\,\mu\text{C}=10^{-5}\,\text{C}$ separated by $2a=5.0\times10^{-3}\,\text{m}$, and the field is wanted at $r=0.15\,\text{m}$ from the centre O. Since $r/a=0.15/0.0025=60\gg1$, the point is far compared with the dipole size, so the short‑dipole approximation is excellent. The dipole moment is

directed from the negative to the positive charge. Take $k=1/4\pi\varepsilon_0=9\times10^{9}\,\text{N·m}^2/\text{C}^2$ and note $r^3=(0.15)^3=3.375\times10^{-3}\,\text{m}^3$.

(a) Point P on the axis. For a short dipole the axial field is

It is directed along the dipole moment (pointing away from the dipole on the positive‑charge side). …

Method: Superposition of Electric Fields (Vector Addition)

This problem uses the principle of superposition: the net electric field at any point is the vector sum of the fields due to each individual charge.

Steps for Part (a) — Point P on the axis

1. Identify the distances

   • Dipole length: $2a = 5.0\ \text{mm} = 5.0 \times 10^{-3}\ \text{m}$
   • Half-length: $a = 2.5 \times 10^{-3}\ \text{m}$
   • Distance from centre O to P: $r = 15\ \text{cm} = 0.15\ \text{m}$
   • Distance from $+q$ to P: $r_+ = r - a = 0.15 - 0.0025 = 0.1475\ \text{m}$
   • Distance from $-q$ to P: $r_- = r + a = 0.15 + 0.0025 = 0.1525\ \text{m}$

2. Write the field magnitudes

   Field due to $+q$ at P:

Field due to $-q$ at P:

3. Add as vectors (same direction) Since both fields point to the right:

4. Substitute values $k = 9 \times 10^9\ \text{N·m}^2/\text{C}^2$, $q = 10\ \mu\text{C} = 10^{-5}\ \text{C}$

Computing:

$\frac{1}{(0.1475)^2} \approx 45.97$, $\frac{1}{(0.1525)^2} \approx 43.00$

Sum $\approx 88.97$

Direction: away from the dipole along the axis, toward the right.

Steps for Part (b) — Point Q on the perpendicular bisector

1. Identify geometry

   • Distance from O to Q: $r = 0.15\ \text{m}$
   • Distance from each charge to Q: $R = \sqrt{r^2 + a^2} = \sqrt{(0.15)^2 + (0.0025)^2}$ Since $a \ll r$, $R \approx r = 0.15\ \text{m}$

2. Field magnitudes are equal

3. Resolve directions

   • $E_+$ points away from $+q$
   • $E_-$ points toward $-q$
   • The horizontal components cancel (equal and opposite)
   • The vertical components add

4. Find vertical component

   From the geometry, $\sin\theta = \frac{a}{R}$

   Each vertical component: $E_{\perp} = \frac{kq}{R^2} \cdot \frac{a}{R} = \frac{kqa}{R^3}$

   Net field (both vertical components add): …

Common Mistakes & How to Avoid Them

Mistake 1: Forgetting the Vector Nature of Electric Field

The Error: Students treat electric field as a scalar and simply add magnitudes without considering direction. For the axial point P, they might compute $E = \frac{kq}{r_1^2} + \frac{kq}{r_2^2}$ directly.

Why It's Wrong: Electric field is a vector. The fields from $+q$ and $-q$ point in opposite directions at P. Simply adding magnitudes gives the wrong answer.

How to Avoid: Always draw arrows showing field directions before calculating. For point P:

• Field from $+q$ points away from $+q$ (to the right)
• Field from $-q$ points toward $-q$ (also to the right)

So they add vectorially: $E_P = E_{+q} + E_{-q}$ (both rightward)

Mistake 2: Using Wrong Distances

The Error: Using $15\,\text{cm}$ as the distance from both charges instead of calculating individual distances.

Why It's Wrong: The charges are $5.0\,\text{mm}$ apart, so distances from P to each charge differ slightly. For P at $15\,\text{cm}$ from centre:

• Distance to $+q$: $r_1 = 15 - 0.25 = 14.75\,\text{cm}$
• Distance to $-q$: $r_2 = 15 + 0.25 = 15.25\,\text{cm}$

How to Avoid: Always compute exact distances from each charge to the point. Draw a clear diagram with labelled distances.

Mistake 3: Unit Conversion Errors

The Error: Mixing cm and mm, or forgetting to convert to metres.

Why It's Wrong: Coulomb's law requires SI units: metres for distance, Coulombs for charge. Using $5.0\,\text{mm}$ as $5.0$ in the formula gives nonsense.

How to Avoid: Convert everything to metres first:

• $5.0\,\text{mm} = 5.0 \times 10^{-3}\,\text{m}$
• $15\,\text{cm} = 0.15\,\text{m}$
• $10\,\mu\text{C} = 10 \times 10^{-6}\,\text{C}$

Mistake 4: Forgetting the Dipole Approximation

The Error: For point Q (equatorial position), students compute exact fields from each charge and add them vectorially, but make sign errors.

Why It's Tricky: At Q, the fields from $+q$ and $-q$ have equal magnitudes but point in different directions. Their horizontal components cancel, vertical components add.

How to Avoid: For the equatorial point:

• Both fields have magnitude $E = \frac{kq}{r^2 + (d/2)^2}$
• The vertical components cancel (opposite directions)
• The horizontal components add: $E_Q = 2E\sin\theta$ where $\sin\theta = \frac{d/2}{\sqrt{r^2 + (d/2)^2}}$

Mistake 5: Using the Wrong Formula for Dipole Field …