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Class 12 › Physics › Electric Potential & Capacitance
CBSE2023Class Class 12 · Physics5 Marks · Short✅ Verified
Derive the expression for the electric potential due to an electric dipole at an axial point.
✅ Answer & Solution
๐ Derivation:
Setup:
โข Dipole: charges +q and โq, separated by distance 2a (dipole length).
โข Dipole moment: <b>p = q ร 2a</b>
โข Point P is on the <b>axial line</b> at distance r from centre.
Step 1 โ Distance from each charge to P:
rโ = r โ a (from +q)
rโ = r + a (from โq)
Step 2 โ Potential at P:
V = k[+q/rโ + (โq)/rโ]
V = kq [1/(rโa) โ 1/(r+a)]
V = kq [(r+a โ r+a) / (rยฒโaยฒ)]
V = kq [2a / (rยฒโaยฒ)]
Step 3 โ In terms of Dipole Moment (p = 2qa):
V = kp / (rยฒ โ aยฒ)
Step 4 โ For Short Dipole (r >> a, so aยฒ << rยฒ):
โโโโโโโโโโโโโโโโโโโโ
โ V = kp / rยฒ โ
โโโโโโโโโโโโโโโโโโโโ
Note: At <b>equatorial point</b>, V = 0 (positive and negative contributions cancel).