#electrostatic equilibrium

Energy Stored in Capacitor and Energy Density

When a conductor is placed in an external electric field, why does the potential throughout its volume become constant i

**Energy density** in electric field u = ½ ε E², ε = K ε₀, E = V/d, total energy U = u·volume = ½ ε E²·A d =½ ε A d·(V/d)²=½ ε A V²/d=½ C V², consistent. For parallel plate, E = V/d ≈10⁶ V/m for 400 V across 0.4 mm, u≈½×8.85×10⁻¹²×10¹²≈4.4 J/m³. In electrostatic equilibrium, the electric field inside a conductor is zero because free charges rearrange to cancel any internal field. Since the electric field is the negative gradient of potential ( E = -(dV/dr) ), if E = 0 , the potential gradient must be zero, implying the potential V is constant

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Energy Stored in Capacitor and Energy Density

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Sharing of Charges, Common Potential and Applications

A \( 14 \, \mu\text{F} \) capacitor charged to \( 20 \, \text{V} \) is connected to an uncharged \( 14 \, \mu\text{F} \)

**Sharing of charges** when charged capacitor C₁ at V₁ connected to uncharged C₂, total charge Q = C₁ V₁ conserved, common potential V_common = Q/(C₁+C₂) = C₁ V₁/(C₁+C₂), final charges Q₁' = C₁ V_common, Q₂' = C₂ V_common. For 4 μF at 100 V (Q=4×10⁻⁴ C) connected to 4 μF uncharged, V_common=4×10⁻⁴/8×10⁻⁶=50 V. Initial charge: Q = 14 × 10⁻⁶ × 20 = 2.8 × 10⁻⁴ C . Total capacitance: 14 + 14 = 28 μF . Final voltage: V = (Q/C) = (2.8 × 10⁻⁴/28 × 10⁻⁶) = 10 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C =

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications

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Equipotential Surfaces and Relation Between Field and Potential

When a charged conductor is placed in contact with an uncharged conductor of smaller size, why does the smaller conducto

**Equipotential surface** is surface where V constant, no work done moving charge along it because W = q ΔV =0 when ΔV=0. Electric field E always perpendicular to equipotential surface, direction from higher to lower potential, magnitude E = -dV/dr, steeper potential gradient means stronger field. When two conductors reach equilibrium, they share charge and attain the same potential. Potential on a conductor's surface is V = (Q/4 π ε₀ R) for a sphere (or similar for other shapes). For equal V , (Q₁/R₁) = (Q₂/R₂) , so Q ∝ R . Surface charge density sigma = (Q/4 π R²) , so sigma ∝ (Q/R²)

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Equipotential Surfaces and Relation Between Field and Potential

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Conductors, Electrostatic Shielding and Dielectrics

Why does the electric field inside a conductor vanish in electrostatic equilibrium, even if the conductor is irregularly

**Electrostatic shielding** inside hollow conducting shell field zero when no charges inside, regardless of external field, because free charges redistribute on outer surface to cancel external field inside conductor, E=0 inside material in equilibrium, consequence of Gauss's law and conductor property. In electrostatic equilibrium, free charges in a conductor redistribute to cancel any internal electric field. If there were a field inside, it would exert a force on the free charges, causing them to move until the field becomes zero everywhere inside. This principle holds regardless of shape b

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Conductors, Electrostatic Shielding and Dielectrics

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Conductors, Electrostatic Shielding and Dielectrics

A metal sphere is placed inside a uniform electric field \( E \). What is the electric field inside the sphere after it

**Electrostatic shielding** inside hollow conducting shell field zero when no charges inside, regardless of external field, because free charges redistribute on outer surface to cancel external field inside conductor, E=0 inside material in equilibrium, consequence of Gauss's law and conductor property. In electrostatic equilibrium, the electric field inside a conductor is zero. When a metal sphere is placed in a uniform electric field, charges redistribute on its surface such that the induced field inside cancels the external field. This results in a net electric field of zero inside the sphe

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Conductors, Electrostatic Shielding and Dielectrics

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Electric Field and Electric Field Lines

Which property of electric charges is responsible for the fact that charges on a conductor in equilibrium reside only on

**Electric field** defined as E = F/q₀, force per unit positive test charge, unit N/C or V/m, direction along force on positive test charge. For point charge, E = k q/r² radially outward for q>0. Field lines start on positive and end on negative, density indicates strength. In electrostatic equilibrium, the electric field inside a conductor must be zero. If charges existed inside, they would create a field, causing further movement. Thus, charges redistribute to the surface, where they can maintain zero internal field due to their mobility. Substituting values gives Mobility, which matches exp

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Field and Electric Field Lines

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Electric Field and Electric Field Lines

What fundamental concept explains why the electric field just outside a charged conductor is perpendicular to its surfac

**Electric field** defined as E = F/q₀, force per unit positive test charge, unit N/C or V/m, direction along force on positive test charge. For point charge, E = k q/r² radially outward for q>0. Field lines start on positive and end on negative, density indicates strength. In electrostatic equilibrium, the electric field inside a conductor is zero. If the field just outside had a tangential component, charges would move along the surface, contradicting equilibrium. Thus, the field must be normal to the surface. Substituting values gives Electrostatic equilibrium, which matches expected magnit

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Field and Electric Field Lines

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