Define electric field and electric potential

1.1.10 · Hardware › Electricity & Charge Basics

Hum in ideas ki zaroorat kyoñ rakhte hain?

Agar charge A aur charge B interact karte hain, toh hum directly unke beech force likh sakte hain (Coulomb's law). Lekin isse dono charges ek doosre se couple ho jaate hain. Physicists chahte the yeh kehna: "charge A akela space ko prepare karta hai; phir jo bhi charge drop karo, bas woh preparation padh lo."

Field ka jawab: "Agar main yahañ ek +1 C test charge rakhun, toh kitni takkat aur kis direction mein dhakka lagega?"
Potential ka jawab: "Infinity se yahañ tak pahunchne mein maine kitna kaam per coulomb kiya?"

Yeh alag karna source vs. test charge ka idea hai, aur yahi poori circuit theory ki neev hai (voltage = potential difference).

Electric Field

$q_{0}$ se divide kyoñ karte hain? Kyoñki hum space ki property chahte hain, visitor ki nahin. Test charge ko double karo toh force bhi double hogi, isliye $F / q_{0}$ constant rehti hai — yeh location ko describe karti hai, charge ko nahin.

Point charge ka field derive karna (scratch se)

Shuru karte hain Coulomb's law se — do point charges ke beech measured force: $F = \frac{1}{4 π ε _{0}} \frac{Q q _{0}}{r ^{2}} \overset{r}{^}$

Ab definition $E = F / q_{0}$ apply karo — test charge ko divide kar do: $E = \frac{1}{4 π ε _{0}} \frac{Q}{r ^{2}} \overset{r}{^}$

Electric Potential

Point charge ka potential derive karna (scratch se)

Work = force times distance. $q_{0}$ ko infinity se $r$ tak laane mein, hum field ke against push karte hain, isliye: $W = - \int_{\infty}^{r} F_{e l ec} \cdot d l = - \int_{\infty}^{r} q_{0} E d r = - \int_{\infty}^{r} q_{0} \frac{k Q}{r ^{2}} d r$

Minus sign kyoñ? Applied work barabar hai minus electric force ke kaam ke (hum usse oppose karte hain).

Integrate karo ( $\int r^{- 2} d r = - r^{- 1}$ ): $W = - q_{0} k Q [- \frac{1}{r}]_{\infty}^{r} = q_{0} k Q (\frac{1}{r} - \frac{1}{\infty}) = \frac{k Q q _{0}}{r}$

$q_{0}$ se divide karo ( $V$ ki definition): $V = \frac{k Q}{r}$

Dono ke beech link

$E = - \frac{d V}{d r} ⟺ V (r) = - \int_{\infty}^{r} E d r$

Figure — Define electric field and electric potential

Worked Examples

Common Mistakes

Active Recall

Recall Khud ko test karo (answers hide karo)

$E$ ki units kya hain? → $N/C = V/m$ .
$V$ ki units kya hain? → $J/C = V$ .
Kya $E$ scalar hai ya vector? Aur $V$ ? → $E$ vector, $V$ scalar.
$E$ aur $V$ mein $q_{0}$ cancel kyoñ hota hai? → Taaki woh space ko describe karein, visiting charge ko nahin.
$E$ aur $V$ ka relationship? → $E = - d V / d r$ ; field potential ka negative slope hai.

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho ek magnet-jaisi ball ne ek pahaadi banayi hai. Potential woh hai kitni unchaayi par tum ho — upar hona matlab stored energy hai. Field woh hai pahaadi kitni steep hai tumhare pair ke neeche — steep matlab bada dhakka. Ball hamesha upar se neeche ki taraf girrti hai, aur sabse zyada tez wahañ jaati hai jahañ pahaadi sabse steep hoti hai. Tumhare phone charger mein voltage sirf "electricity ki pahaadi kitni uchi hai" hai, aur yahi cheez tiny charges ko wire ke through push karti hai.

Connections

Coulomb's Law — woh force law jise hum ne $q_{0}$ se divide karke dono quantities nikaalin.
Electric Potential Energy — $U = q_{0} V$ , energy normalise karne se pehle.
Voltage and Potential Difference — circuits mein practical use ( $V_{i} - V_{f}$ ).
Equipotential Surfaces and Field Lines — field lines ⟂ equipotentials.
Capacitance — plates ke beech field aur potential difference se seedha banta hai.
Gauss's Law — symmetric charge ke liye $E$ nikalane ka tez raasta.

Electric field definition

Unit positive test charge par force per unit,

E = F / q_{0}

; N/C mein ek vector.

Electric potential definition

Positive charge ko infinity se laane mein work done per unit charge,

V = W / q_{0}

; volts (J/C) mein ek scalar.

Field of a point charge

E = k Q / r^{2}

, direction + se door / − ki taraf.

Potential of a point charge

V = k Q / r

(scalar, koi direction nahin).

Why does field fall as 1/r² but potential as 1/r?

Potential field ka distance par integral hai;

\int r^{- 2} d r \propto r^{- 1}

Relation between E and V

E = - d V / d r

; field potential ka negative slope (gradient) hai.

Units of E and V

E

: N/C = V/m;

V

: J/C = volt.

Can E=0 while V≠0?

Haan — jaise hollow charged conductor ke andar.

Can V=0 while E≠0?

Haan — jaise equal + aur − charges ke beech midpoint par.

Force on charge q in field E

F = q E

; negative q ke liye direction ulti ho jaati hai.

Work by field moving q between potentials

W = q (V_{i} - V_{f})

Value of Coulomb constant k

k = 1/ (4 π ε_{0}) \approx 8.99 \times 1 0^{9} N\cdotm^{2} / C^{2}