NCERT Plus Two (Class 12) Physics • Chapter 2

Electrostatic Potential and Capacitance

Q: Define an equipotential surface. Why can two equipotential surfaces never intersect?

An equipotential surface is a surface over which electrostatic potential V has the same constant value at every point. No intersection: If two equipotential surfaces crossed, the intersection point would simultaneously have two different potential values (and two directions of E ), which is physically impossible.

Q: Derive the expression for electrostatic potential at a point due to an electric dipole at distance r ≫ a and angle θ. Hence deduce equatorial potential.

Let P be at distance r₁ from +q and r₂ from −q (separation 2a ). V = kq (1/r₁ − 1/r₂) . Using geometry for r ≫ a : r₁ ≈ r − a cosθ , r₂ ≈ r + a cosθ . 1/r₁ − 1/r₂ ≈ 2a cosθ / r² ⇒ V = p cosθ / (4πε₀ r²) (since p = 2qa ). On equatorial plane ( θ = 90° ): cos 90° = 0 ⇒ V = 0 .

Q: What is electrostatic shielding? Give one practical application.

Electrostatic shielding is the phenomenon of making a cavity inside a conductor completely free from external electric fields ( E inside = 0 ), regardless of external charges or fields. Application: Staying inside a closed metal car protects passengers from lightning strikes (or coaxial cables shield signals).

Q: A parallel-plate capacitor with air is charged by a battery to potential V₀. Keeping the battery connected, a dielectric slab (K) fills the gap. How do (i) Capacitance, (ii) Charge, (iii) Electric field, and (iv) Stored energy change?

Since battery stays connected, V = V₀ remains constant : (i) Capacitance C = K C₀ (increases K times). (ii) Charge Q = CV = K Q₀ (increases K times). (iii) Electric field E = V/d = E₀ (unchanged). (iv) Energy U = ½CV² = K U₀ (increases K times).

Q: Derive the expression for capacitance of a parallel-plate capacitor filled with a dielectric slab of thickness t (< d) and dielectric constant K.

Air gap thickness = (d − t) with field E₀ = σ/ε₀ ; dielectric thickness = t with field E = E₀/K = σ/(Kε₀) . Total potential difference: V = E₀(d − t) + (E₀/K)t = Q ε₀A (d − t + t K ) . Hence C = Q/V = ε₀ A / (d − t + t/K) .

Q: Derive an expression for the energy stored in a charged capacitor. Why is half the energy supplied by the battery lost during charging?

Work to move small charge dq at potential V' = q/C is dW = (q/C) dq . Integrating from 0 to Q : U = ∫₀^Q (q/C) dq = Q²/(2C) = ½ CV² = ½ QV . The battery supplies total work W batt = QV ; the remaining ½ QV dissipates as heat in circuit resistance during charging.

Q: A 12 pF capacitor is connected to a 50 V battery. How much energy is stored? If it is disconnected and connected across another uncharged 12 pF capacitor, find the new stored energy.

Initial energy U₁ = ½ C₁ V² = ½ (12×10⁻¹²)(50²) = 1.5 × 10⁻⁸ J . Sharing charge equally between two identical 12 pF capacitors halves voltage to V' = 25 V across C eq = 24 pF : U₂ = ½ (24×10⁻¹²)(25²) = 0.75 × 10⁻⁸ J (50% energy lost as spark/heat!).

Q: Two point charges +5 × 10⁻⁸ C and −3 × 10⁻⁸ C are 16 cm apart. At what point(s) on the line joining them is the electric potential zero?

Case 1 (between charges, distance x from +5): 5/x − 3/(16 − x) = 0 ⇒ 5(16 − x) = 3x ⇒ 8x = 80 ⇒ x = 10 cm from +5 . Case 2 (outside, beyond −3 at distance y from +5): 5/y − 3/(y − 16) = 0 ⇒ 5y − 80 = 3y ⇒ 2y = 80 ⇒ y = 40 cm from +5 (i.e. 24 cm beyond −3 ).

Point & dipole potentials, equipotential surfaces (E = −dV/dr), system potential energy, conductors & dielectrics, parallel-plate capacitors & stored energy.

📄 Official NCERT Textbook PDF

Topic 2.2 – 2.4: Electrostatic Potential due to Point Charge & Electric Dipole

Carrying a heavy schoolbag up the steep stairs of Munnar takes work — and the higher you climb, the more gravitational height energy each kilogram stores! Similarly, pushing a tiny +1 C charge from far away up close to a positive charge builds an electrical height called Electrostatic Potential.

POINT CHARGE VV = W_∞→Pq = 14πε₀ Qr (Scalar! Falls as 1/r; Unit: Volt = J/C)

DIPOLE V (r ≫ a)V = 14πε₀ p cosθr² (Falls faster as 1/r²!)

EQUATORIAL ZEROAt θ = 90°: V = 0 everywhere, yet E ≠ 0!

🧠 Memory Trick: Watch the r powers drop cleanly: Point V ∝ 1/r (E ∝ 1/r²), Dipole V ∝ 1/r² (E ∝ 1/r³)! And V is a SCALAR (keep + or − signs!).

💡 Simple Kid-Check: Force and Electric Field are arrows (vectors) that push in a direction, which makes adding them tricky. Potential V is just a single temperature-like number (scalar) in Volts — positive near +Q, negative near −Q!

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Definition of Potential: Work done per unit positive test charge in moving it slowly (zero acceleration) from infinity to point P against electrostatic force: V = W_ext / q = kQ / r. Only potential difference ΔV = V_B − V_A = W_AB/q is physically measurable (path-independent, because electrostatic field is conservative!).
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Dipole Potential Formula: Superposing +q and −q separated by 2a at far point (r, θ) gives V = 14πε₀ p cosθr² = p · r̂4πε₀r².
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Special Dipole Positions: On axis (θ = 0° or 180°): V = ±kp / r². On perpendicular bisector (θ = 90°): cos 90° = 0 ⇒ V = 0 (equal distances from +q and −q cancel!).

🧒 Super-Easy Example: Find potential 10 cm from a +4 μC charge: V = (9 × 10⁹)(4 × 10⁻⁶) / 0.10 = 3.6 × 10⁵ V. Moving a 2 nC charge from infinity to this spot takes work W = qV = 7.2 × 10⁻⁴ J!

Top Kerala HSE 1-mark & 2-mark trap: Can Electric Potential be zero at a point where Electric Field is NON-ZERO? YES! On the equatorial line of an electric dipole (or midway between +q and −q), V = +kq/r − kq/r = 0, yet electric field E = kp/r³ is definitely non-zero! Conversely, inside a charged hollow metal shell, E = 0 while V = kQ/R ≠ 0!

Topic 2.5 – 2.6: System Potential & Equipotential Surfaces (E = −dV/dr)

Look at a hiker's contour map of Wayanad hills: every point along one contour ring sits at the exact same altitude above sea level, so walking sideways along a ring needs ZERO climbing effort! In Physics, those constant-voltage contour rings are Equipotential Surfaces.

WORK = ZERO!On any equipotential surface, ΔV = 0 ⇒ W = q ΔV = 0!

FIELD GRADIENTE = −dVdr (E always points toward decreasing V & is ⊥ to surface!)

SURFACE SHAPESPoint charge: Concentric spheres (crowded near charge!) • Uniform E: Parallel planes

🧠 Memory Trick: dr = dV / E — for equal voltage steps dV, equipotentials pack CLOSER where E is strong and spread apart where E is weak!

💡 Simple Kid-Check: Just as water always rolls downhill perpendicular to map contour lines, a positive charge pushed by E always accelerates toward lower potential V, cutting every equipotential surface at an exact 90° angle!

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Potential of Many Charges: Scalar superposition makes it super easy: V_net = 14πε₀ Σ qᵢrᵢ. Note: on the bisecting plane of a dipole, or halfway between +q and −q, V = 0!
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3 Must-Know Properties of Equipotential Surfaces:
(a) Zero Work: Moving a charge anywhere along an equipotential requires W = q(V_B − V_A) = 0.
(b) Always Perpendicular to E: Because dW = qE · dr = 0 ⇒ E dr cosθ = 0 ⇒ θ = 90°.
(c) Never Intersect: Crossing surfaces would force two different potentials at one spot — impossible!
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Connecting E and V: E = −dV/dr (explaining why N/C ≡ V/m!). The minus sign means electric field always points in the direction where potential decreases steepest!

🧒 Super-Easy Example: Two parallel planes at x = 0 (V = 60 V) and x = 2 cm (V = 20 V) give uniform field E = −(20 − 60)/0.02 = +2000 V/m pointing along +x (from high 60 V down to low 20 V!).

Watch three board-exam gems: (1) Why are equipotential spheres around a point charge NOT equally spaced for equal ΔV? Because E ∝ 1/r² weakens outward, so Δr = ΔV/E grows wider as r increases! (2) Moving a charge from A to B on a spherical equipotential of radius R around +Q always costs ZERO work, even if you take a crazy curved detour! (3) Don't drop the minus sign in E = −dV/dr!

Topic 2.7 – 2.8: Potential Energy of Charge Systems & Dipole in Uniform Field

Imagine squeezing two stiff springs together with your palms — you feel energy packed inside waiting to bounce back! Bringing two positive charges close together from infinity stores that exact same spring-like Electrostatic Potential Energy U inside the pair!

2-CHARGE PAIR UU = 14πε₀ q₁ q₂r₁₂ (Unit: Joule or eV; 1 eV = 1.6×10⁻¹⁹ J)

3-CHARGE SYSTEMSum all 3 pairs: U = k(q₁q₂r₁₂ + q₂q₃r₂₃ + q₁q₃r₁₃)

DIPOLE ENERGYU(θ) = −pE cosθ = −p·E • Rotation work: W = pE(cosθ₁ − cosθ₂)

🧠 Memory Trick: For Like charges (q₁q₂ > 0), pushing them closer raises U > 0; for Unlike charges (q₁q₂ < 0), they attract so U < 0! For dipole rotation from 0° → 180°, work = 2pE!

💡 Simple Kid-Check: Think of building a 3-charge triangle one brick at a time: bringing charge 1 into empty space costs ZERO work; charge 2 feels charge 1; charge 3 feels BOTH 1 and 2 — giving three pair terms in total!

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Assembling Charges in Empty Space: First charge q₁ enters empty space freely (W₁ = 0). Bringing q₂ at distance r₁₂ costs W₂ = q₂V₁ = k q₁q₂ / r₁₂. For n charges, sum over every unique pair (i < j).
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Charges in an External Field V(r): Total energy adds external potential work to mutual pair interaction: U = q₁V(r₁) + q₂V(r₂) + kq₁q₂r₁₂!
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Dipole Potential Energy: Choosing zero energy at θ = 90°, U(θ) = −pE cosθ.
• θ = 0° (aligned): U_min = −pE (Stable!).
• θ = 90°: U = 0.
• θ = 180° (flipped): U_max = +pE (Unstable! Work from 0° → 180° = (+pE) − (−pE) = 2pE!).

🧒 Super-Easy Example: Two protons move closer from distance 2r to r. Since both are positive (+), repulsion fights you → Potential Energy INCREASES (doubles)! What about an electron & proton moved closer? Since U = −ke²/r is negative, moving closer makes U MORE negative → Potential Energy DECREASES!

Watch three favourite exam traps: (1) Students forget to include the external potential terms q₁V(r₁) + q₂V(r₂) when assembling two charges inside an external field! (2) Rotating a dipole from stable equilibrium (0°) by 90° costs W = pE, whereas flipping 180° costs 2pE (NOT zero!). (3) Know 1 eV = 1.6 × 10⁻¹⁹ J!

Topic 2.9 – 2.10: Electrostatics of Conductors, Shielding & Dielectrics

Why do pilots and drivers stay safely inside a closed metal car during a fierce Kerala monsoon thunderstorm? Because a metal body acts like an invisible Faraday cage — free electrons rearrange instantly on the outside skin so the electric field inside stays ZERO!

CONDUCTOR IN EQUILIBRIUMInside bulk & cavity: E = 0 • V = constant • Excess charge sits ONLY on outer surface!

CONDUCTOR SURFACEField is strictly normal to surface: E = σε₀ n̂

DIELECTRICSNon-conductors polarise! Net field shrinks to E = E₀K | P = χₑ ε₀ E

🧠 Memory Trick: Conductors CANCEL field inside completely (E_in = 0, Faraday shielding!); Dielectrics REDUCE field inside by factor K (E_in = E₀/K)!

💡 Simple Kid-Check: In a conductor (metal), electrons are free athletes that sprint to the walls until they cancel every trace of external electric field inside (E = 0). In a dielectric (glass, plastic, mica), electrons are tied to atoms on tiny springs — they just stretch slightly (polarisation) to weaken the field!

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5 Electrostatic Rules of a Conductor: (1) E = 0 inside; (2) Net charge inside is 0 (lives only on surface); (3) E at surface is normal σ/ε₀; (4) Potential V is constant throughout volume & surface; (5) Electrostatic Shielding: even if a conductor carries huge external charge or sits in a strong external field, E = 0 inside any hollow cavity!
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Polar vs Non-Polar Dielectrics:
• Non-polar (H₂, O₂, CO₂): centers of + and − coincide until external field stretches them.
• Polar (H₂O, HCl): permanent built-in dipole moment, randomly jumbled by heat until external E₀ aligns them!
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Polarisation Vector P: Induced dipole moment per unit volume: P = χₑ ε₀ E. Bound surface charge σ_p sets up opposing field E_p, cutting internal field to E = E₀ − E_p = E₀ / K!

🧒 Real-Life Protection: Sensitive electronic cables (coaxial cables) and delicate lab instruments sit inside braided copper shields — external lightning or stray electric noise induces surface charges on the outer metal shell, leaving the inside dead calm (E = 0)!

Two classic DHSE traps: (1) Why is it safer inside a closed car than under a tree during lightning? Because of Electrostatic Shielding (E = 0 inside hollow conductor cavity, NOT because rubber tyres insulate lightning!). (2) In a dielectric, do NOT confuse polarisation charges (bound to surface, cannot flow) with free conductor charges — inside a dielectric, net field is reduced to E₀/K, NOT zero!

Topic 2.11 – 2.13: Capacitors, Parallel-Plate Formula & Dielectric Slab

Need a sudden lightning-bolt burst of current for your phone's camera flash or to kick-start a Kerala ceiling fan? A battery releases energy slowly, like filling a water tank with a tap — a Capacitor acts like a quick splash bucket that dumps stored charge in milliseconds!

CAPACITANCEC = QV (Unit: Farad F = C/V; depends ONLY on geometry & medium!)

PARALLEL-PLATEAir: C₀ = ε₀ Ad | Full Dielectric: C = K ε₀ Ad = K C₀

PARTIAL SLAB (t)C = ε₀ Ad − t + t/K (Metal slab K = ∞ ⇒ t/K = 0!)

🧠 Memory Trick (Dielectric Insertion): Battery CONNECTED ⇒ V fixed (Q↑ K×, U↑ K×); Battery DISCONNECTED ⇒ Q fixed (V↓ 1/K, E↓ 1/K, U↓ 1/K)! Capacitance C ALWAYS grows K×!

💡 Simple Kid-Check: Capacitance is like a water bucket's width — doubling how much charge Q you pour in doubles the voltage height V, so the ratio C = Q/V stays FIXED by the plates' area A, gap d, and filling K!

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Parallel-Plate Derivation (3 Lines!): Two plates of area A carry +σ and −σ (σ = Q/A). Between plates, fields add: E = σ/ε₀ = Q/(ε₀A). Potential difference V = E·d = Qd/(ε₀A) ⇒ C₀ = Q/V = ε₀ A / d!
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Why Dielectrics Boost Capacitance: Sliding in a dielectric reduces internal field E → E₀/K, which drops voltage V → V₀/K for the same charge, shooting capacitance up C = K C₀!

#1 Board Exam Comparison Table (Insert Dielectric K):

Quantity	Battery Connected (`V` const)	Battery Disconnected (`Q` const)
Capacitance `C`	Increases `K` times (`KC₀`)	Increases `K` times (`KC₀`)
Charge `Q`	Increases `K` times (`KQ₀`)	Constant (`Q₀`)
Voltage `V`	Constant (`V₀`)	Decreases `1/K` (`V₀/K`)
Electric Field `E = V/d`	Constant (`E₀`)	Decreases `1/K` (`E₀/K`)
Energy `U`	Increases `K` times (`KU₀`)	Decreases `1/K` (`U₀/K`)

Top Kerala HSE 2/3-mark trap: Does capacitance C = Q/V increase if you double the charge Q? NO! Doubling Q automatically doubles V, leaving C constant! Also ALWAYS check whether the battery stays connected or is disconnected before sliding a dielectric slab — Q, V, and Energy U flip between multiplying by K and dividing by K!

Topic 2.14 – 2.15: Series / Parallel Capacitors & Energy Stored (½ CV²)

Think about connecting two identical water buckets side-by-side (Parallel) versus stacking one narrow pipe below another (Series)! Parallel gives double total width (capacitances ADD!), while Series shares the same charge trickle across split voltages (capacitances shrink like parallel resistors!).

SERIES (Same Q)1C_eq = 1C₁ + 1C₂ | V = V₁ + V₂ (C_eq < smallest C!)

PARALLEL (Same V)C_eq = C₁ + C₂ | Q = Q₁ + Q₂ (C_eq > largest C!)

STORED ENERGYU = 12CV² = 12QV = Q²2C | Energy Density: u = 12ε₀E²

🧠 Memory Trick: Capacitors combine OPPOSITE to Resistors! Parallel capacitors ADD (C₁ + C₂); Series capacitors take reciprocals (1/C₁ + 1/C₂)! And battery sends QV energy while capacitor stores only ½ QV!

💡 Simple Kid-Check: Why does a capacitor only store ½ QV instead of QV? Because at the start of charging, the plate voltage is 0 V, rising linearly to V — so the average voltage you push charge against is ½ V! The other ½ QV heats up the wire!

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Series Combination: Connected end-to-end ⇒ SAME charge Q sits on every capacitor, while voltage divides inversely with capacitance (Vᵢ = Q/Cᵢ): 1C_s = 1C₁ + 1C₂ + .... For 2 capacitors: C_s = C₁C₂C₁+C₂.
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Parallel Combination: Connected across same two nodes ⇒ SAME voltage V across every branch, while charges add (Qᵢ = CᵢV): C_p = C₁ + C₂ + ....
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Energy & Energy Density: Integrating dW = (q/C) dq gives U = Q²2C = 12CV². Dividing by plate volume A·d yields electrostatic energy density u = ½ ε₀ E² (energy actually lives in the electric field between plates!).

🧒 Super-Easy Example: Two 6 μF capacitors are connected across 12 V.
• In Series: C_s = 6/2 = 3 μF ⇒ U = ½(3 μF)(12²) = 216 μJ (each gets 6 V!).
• In Parallel: C_p = 6 + 6 = 12 μF ⇒ U = ½(12 μF)(12²) = 864 μJ (4× bigger!).

Watch two major DHSE traps: (1) Capacitors combine backwards from resistors — don't accidentally add series capacitors directly! (2) When a charged capacitor C₁ at V₁ is connected in parallel to an uncharged capacitor C₂, total charge is conserved (V_common = C₁V₁/(C₁+C₂)), yet total energy DECREASES because half/some energy is lost as heat and spark in the connecting wire!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 2 Marks): Define an equipotential surface. Why can two equipotential surfaces never intersect?

An equipotential surface is a surface over which electrostatic potential V has the same constant value at every point.
No intersection: If two equipotential surfaces crossed, the intersection point would simultaneously have two different potential values (and two directions of E), which is physically impossible.

Q2 (Kerala HSE Say • 3 Marks): Derive the expression for electrostatic potential at a point due to an electric dipole at distance r ≫ a and angle θ. Hence deduce equatorial potential.

Let P be at distance r₁ from +q and r₂ from −q (separation 2a).
V = kq (1/r₁ − 1/r₂). Using geometry for r ≫ a: r₁ ≈ r − a cosθ, r₂ ≈ r + a cosθ.
1/r₁ − 1/r₂ ≈ 2a cosθ / r² ⇒ V = p cosθ / (4πε₀ r²) (since p = 2qa).
On equatorial plane (θ = 90°): cos 90° = 0 ⇒ V = 0.

Q3 (Kerala HSE March • 2 Marks): What is electrostatic shielding? Give one practical application.

Electrostatic shielding is the phenomenon of making a cavity inside a conductor completely free from external electric fields (E_inside = 0), regardless of external charges or fields.
Application: Staying inside a closed metal car protects passengers from lightning strikes (or coaxial cables shield signals).

Q4 (Kerala HSE Model • 2 Marks): A parallel-plate capacitor with air is charged by a battery to potential V₀. Keeping the battery connected, a dielectric slab (K) fills the gap. How do (i) Capacitance, (ii) Charge, (iii) Electric field, and (iv) Stored energy change?

Since battery stays connected, V = V₀ remains constant:
(i) Capacitance C = K C₀ (increases K times).
(ii) Charge Q = CV = K Q₀ (increases K times).
(iii) Electric field E = V/d = E₀ (unchanged).
(iv) Energy U = ½CV² = K U₀ (increases K times).

Q5 (Kerala HSE March • 3 Marks): Derive the expression for capacitance of a parallel-plate capacitor filled with a dielectric slab of thickness t (< d) and dielectric constant K.

Air gap thickness = (d − t) with field E₀ = σ/ε₀; dielectric thickness = t with field E = E₀/K = σ/(Kε₀).
Total potential difference: V = E₀(d − t) + (E₀/K)t = Qε₀A (d − t + tK).
Hence C = Q/V = ε₀ A / (d − t + t/K).

Q6 (Kerala HSE Say • 2 Marks): Derive an expression for the energy stored in a charged capacitor. Why is half the energy supplied by the battery lost during charging?

Work to move small charge dq at potential V' = q/C is dW = (q/C) dq.
Integrating from 0 to Q: U = ∫₀^Q (q/C) dq = Q²/(2C) = ½ CV² = ½ QV.
The battery supplies total work W_batt = QV; the remaining ½ QV dissipates as heat in circuit resistance during charging.

Q7 (Kerala HSE March • 3 Marks): A 12 pF capacitor is connected to a 50 V battery. How much energy is stored? If it is disconnected and connected across another uncharged 12 pF capacitor, find the new stored energy.

Initial energy U₁ = ½ C₁ V² = ½ (12×10⁻¹²)(50²) = 1.5 × 10⁻⁸ J.
Sharing charge equally between two identical 12 pF capacitors halves voltage to V' = 25 V across C_eq = 24 pF:
U₂ = ½ (24×10⁻¹²)(25²) = 0.75 × 10⁻⁸ J (50% energy lost as spark/heat!).

Q8 (NCERT / DHSE Numerical • 2 Marks): Two point charges +5 × 10⁻⁸ C and −3 × 10⁻⁸ C are 16 cm apart. At what point(s) on the line joining them is the electric potential zero?

Case 1 (between charges, distance x from +5): 5/x − 3/(16 − x) = 0 ⇒ 5(16 − x) = 3x ⇒ 8x = 80 ⇒ x = 10 cm from +5.
Case 2 (outside, beyond −3 at distance y from +5): 5/y − 3/(y − 16) = 0 ⇒ 5y − 80 = 3y ⇒ 2y = 80 ⇒ y = 40 cm from +5 (i.e. 24 cm beyond −3).