WHY define a "per unit charge" quantity? Because the force on a charge scales with q0, so the
work also scales with q0. Dividing it out gives a property of the field/location alone —
something you can map once and reuse for any charge.
Q: A +2μC charge moves from a point at V=50 V to a point at V=20 V. Does the field do
positive or negative work, and how much?
Forecast first, then open:
Work by field =q(Va−Vb)=(2×10−6)(50−20)=+6×10−5 J. Positive —
the charge moved "downhill" in potential, just like a ball rolling down.
Work done per unit positive charge to bring it from the reference (∞) to that point, without changing KE. Units: volts (J/C).
State the defining integral relation between V and E.
Vb−Va=−∫abE⋅dl.
Why the minus sign in V=−∫E⋅dl?
Because the integral is work done by the field; potential decreases in the direction E points (field points "downhill").
Is electric potential scalar or vector?
Scalar — it has magnitude only; direction info is in E=−∇V.
Derive V(r) for a point charge.
V=−∫∞rr′2kqdr′=kq/r.
How do you get E from V?
E=−∇V (negative gradient of the scalar potential).
For a uniform field, relate E, ΔV and plate separation d.
E=ΔV/d, with E pointing from high to low potential.
Work done by the field moving charge q from a to b?
Wfield=q(Va−Vb).
When does choosing V=0 at infinity fail?
For infinite charge distributions (line/plane) — the integral diverges; use a finite reference.
Recall Feynman: explain to a 12-year-old
Think of a skateboard ramp. Potential is just how high up the ramp you are — higher means more
"stored go". Electric field is how steep the ramp is, and it always points downhill.
If you stand still anywhere, the height tells you how much energy you can release. The formula
V=−∫E⋅dl literally says: "to find the height, add up the steepness as you walk,
and put a minus because walking the way the field pushes means going down."
Dekho, electric potential ka basic idea ek pahaadi (hill) jaisa samajh lo. Pahaadi ki height
batati hai ki tum kitni energy store kiye baithe ho — wahi cheez electric world me potentialV
hai, lekin "per unit charge". Aur electric field E us pahaadi ka dhalaan (slope) hai,
jo hamesha neeche ki taraf, yaani decreasing potential ki direction me point karta hai. Isiliye formula
me minus aata hai: V=−∫E⋅dl.
Derivation simple hai. Field charge par force lagata hai F=q0E. Agar tum charge
ko bina speed badle slowly le jaate ho, to bahar se utna hi opposite force lagana padta hai, aur uska
kiya gaya kaam = potential energy ban jaata hai. Kaam ko q0 se divide karo to potential difference
mil jaata hai. Bas, Vb−Va=−∫abE⋅dl ready hai. Point charge ke liye
radial path leke karoge to seedha V=kq/r aa jaata hai — ye sabse important result hai.
Ek aur cheez yaad rakho: V ek scalar hai (sirf number), E vector hai. Inverse relation
E=−∇V — yaani pehle aasan scalar V nikalo, phir gradient lekar mushkil vector
E paa lo. Exam me ye trick 80/20 waali hai, time bachata hai. Sign ko leke ghabrao mat:
positive charge ke paas V bada hota hai, aur field bahar ki taraf (lower V) point karta hai —
ye consistency check har baar use karo.
