⚡️ Electrostatic Potential – Quick Guide

The electrostatic potential V at a point tells us how much work we must do to gently bring a unit + charge from very far away (infinity) to that spot, without giving it any extra speed 🔋. The work is done against the electrostatic force, and it doesn’t matter which route we take — only the starting and ending points count! :contentReference[oaicite:0]{index=0}

🎯 Potential Created by a Single Point Charge

  • Place a charge Q at the origin. At distance r from it, the potential is \( V(r) \;=\; \dfrac{Q}{4\pi \varepsilon_0\, r} \) 🌟 :contentReference[oaicite:1]{index=1}
  • Sign matters: • For Q > 0, V is positive (you push against repulsion). • For Q < 0, V is negative (attraction does the work for you 😉). :contentReference[oaicite:2]{index=2}
  • We choose V = 0 at infinity so the formula above automatically satisfies this reference. :contentReference[oaicite:3]{index=3}
  • How we got the formula (idea only): – Force on the test charge at an intermediate point is \( \dfrac{1}{4\pi \varepsilon_0}\, \dfrac{Q}{r’^2}\,\hat r’ \). – Add up tiny bits of work \( dW = -\,\dfrac{Q}{4\pi \varepsilon_0}\dfrac{dr’}{r’^2} \) from \( r’ = \infty \) to \( r’ = r \). – The integral gives the neat \( 1/r \) expression above. :contentReference[oaicite:4]{index=4}

📈 How Potential and Field Fall Off with Distance

A point charge’s potential drops as \( 1/r \), while its electric field falls faster, as \( 1/r^2 \). The graph below (conceptual) shows both curves: the blue line (potential) declines gently; the black line (field) dives more steeply. :contentReference[oaicite:5]{index=5}

🔄 Path Independence

No matter how twisty your path, the work done in moving a charge between two fixed points in an electrostatic field stays the same. Break any weird route into tiny radial and sideways hops: the sideways parts add zero work, leaving only the simple radial contribution 🚀. :contentReference[oaicite:6]{index=6}

📝 Worked Example (Try It!)

  1. Potential at P: A \( 4 \times 10^{-7}\,\text{C} \) charge sits \( 9 \,\text{cm} \) away. \( V = \dfrac{4 \times 10^{-7}}{4\pi \varepsilon_0 \times 0.09} \approx 4 \times 10^{4}\,\text{V} \).
  2. Bringing another charge in: For \( q = 2 \times 10^{-9}\,\text{C} \), work done is \( W = qV = 2 \times 10^{-9}\,\text{C} \times 4 \times 10^{4}\,\text{V} = 8 \times 10^{-5}\,\text{J} \). Again, the answer is the same along any path 👍. :contentReference[oaicite:7]{index=7}

🎯 High-Yield Ideas for NEET

  • Must-know formula: \( V = \dfrac{Q}{4\pi\varepsilon_0 r} \).
  • Sign of potential mirrors the sign of charge (+ gives +V, – gives –V).
  • Work done & potential are path independent in electrostatics.
  • Zero potential chosen at infinity is a handy reference in problems.
  • The \( 1/r \) vs \( 1/r^2 \) drop-off helps link potential to electric field quickly.

Keep these nuggets in mind, practice a few numeric questions, and you’ll ace related NEET problems with confidence! ✨