Electric Potential & Energy of Systems Electric Potential & Energy of Systems Electric potential is a simple, powerful idea: it measures how much electric potential energy per unit charge a point in space has. If a tiny positive test charge q is placed at a point, the energy associated with that placement is U = qV ; so V is the energy tag that the location carries, independent of which test charge you choose. Because V is a scalar, potentials from multiple charges add up algebraically; that makes potential calculations far easier than vector sums of fields. Intuitively, this is just like gravity: a high place has higher gravitational potential; a positive charge creates a “high” electric potential around it, and a negative charge creates a “low” potential. Moving a charge from one point to another changes its potential energy by U = q , V . In electrostatics, forces derive from a potential, so the electric field is related to potential by the gradient: E = - V . Along one dimension (radial direction around an isolated charge), this reduces to E = - dV dr . Equipotential surfaces are the contour maps of V : if you slide a charge along an equipotential, its potential energy does not change and the electric field does no work. For a system of point charges, the total potential at a point is the sum V = i k ,q i/r i , and the potential energy stored in the configuration is the sum over pairs U = i<j k ,q i q j / r ij . Negative potential energy signals a bound arrangement (like unlike charges at finite separation); positive U signals a configuration that “wants” to fly apart unless work is done to hold it. These ideas let you answer nearly every NEET problem on work, energy, and movement of charges without messy vector algebra—by tracking scalars carefully and respecting sign conventions and reference levels (usually V = 0 at infinity). Gravity analogy: Climbing a hill raises gravitational potential. Similarly, moving a positive charge toward a positive source raises electric potential (you must do work), and moving it toward a negative source lowers electric potential (the field does work). remember Electric Potential (V) Energy per unit charge at a point: V = U/q . Scalar; reference usually set as V( )=0 . Electric Potential Energy (U) Energy of a charge or system of charges because of configuration in an electric field. For a test charge: U = qV . Reference at Infinity Choice V( )=0 in electrostatics; work to bring a charge from infinity to a point defines potential. Equipotential Surface Set of points with the same potential. No work is done moving a charge along it; E is perpendicular to it. A field derivable from a potential. In electrostatics, E d l =0 and E =- V . Conservative Field Spatial rate of change of potential. Electric field is minus the gradient: E =- V . Potential Gradient Total potential energy of a system of charges: sum over all distinct pairs i<j . Configuration Energy Bound State A configuration with negative potential energy ( U<0 ), requiring energy input to separate to infinity. From Work to Potential: Definitions That Matter Potential difference via work Potential drop equals negative line integral of field along any path (path independent in electrostatics). This definition mirrors gravity: if you walk with the electric field, potential decreases; against it, potential increases. Because the electrostatic field is conservative, the integral depends only on the endpoints, not the path. Setting V( )=0 gives absolute potential: V(P) = - P E d l . Field–potential link The electric field points downhill on the potential landscape. For spherically symmetric systems, this equation allows determining the radial electric field directly from potential. When symmetry is spherical (point charge, spherical shell), use E r = - dV dr . In Cartesian problems without symmetry, use components: E x = - V/ x , etc. tip Electrostatics (charges stationary). Reference potential V( )=0 . Point charge q in free space. V(r) = 1 4 0 q r Field of a point charge is inverse-square and radial. Choose the straight radial path; E d l . V of a point charge Potential at distance r from a point charge q (zero at infinity) is V = k q / r; sign follows q. q = +2.0 , C = 2.0 10 -6 , C r = 0.20 , m k = 8.99 10 9 , N ,m 2 /C 2 V at r = 0.20 m easy Find the electric potential at a point 20 cm away from a +2.0 µC point charge. Take vacuum. A positive q gives positive V ; a negative q gives negative V . Magnitude falls as 1/r . At r , V 0 . Close to the charge, |V| becomes large. Superposition of Potentials and Equipotential Surfaces Because potential is a scalar, it superposes simply: for charges q 1, q 2, , V(P) = i k ,q i/r i . This is often much easier than adding E vectors. Equipotential surfaces are the contour lines of this scalar field. Moving a charge along an equipotential costs zero work ( U = q , V = 0 ), so the electric field must be perpendicular to every equipotential surface. Add potentials directly; keep charge signs and distances straight. Superposition of V Common shapes: Around a single point charge, equipotentials are concentric spheres. Between parallel plates, they are planes parallel to the plates (uniform spacing means uniform field). Around a dipole, equipotentials are closed surfaces that pinch near the charges and reflect the dependence. neet-alert No work along an equipotential: W = 0. A frequent trap is to compute work using W = qEd even if the path is at an angle. If start and end are on the same equipotential, V = 0 so W = q , V = 0 , regardless of path. Single point charge Concentric spheres (V ∝ 1/r) Radial; outward for +q, inward for −q Two like charges Bulging lobes; high V region between charges Field lines curve outward away from midline Dipole (+q, −q) Hourglass-like surfaces; V changes sign across equatorial plane From +q to −q; tangent to lines perpendicular to equipotentials Parallel plates (uniform field) Parallel planes, equally spaced Uniform, from high V plate to low V plate Charge setup Equipotential surfaces Electric field direction Distance r from point charge V=k q / R r=R Potential at finite R Reference level V→0 r→∞ Potential V Large to 0 Potential of a point charge vs distance: 1/r dependence with sign of q. control derived V(r)=k q / r A rectangular hyperbola decreasing from large magnitude near r≈0 to 0 as r→∞. For q>0, V>0; for q<0, V<0. custom Potential Difference and Work–Energy Potential difference is the energy change per unit charge. If a charge q moves from A to B, the change in its potential energy is U = q ,[V(B)-V(A)] . The work done by the electric field is W field = - U , and the work you must do externally (slowly, no kinetic energy change) is W ext = + U . This approximation is valid when the observation point (r) is much greater than the separation distance of the dipole charges (r d). U and V link Scalar bookkeeping: watch the sign of q and the direction of potential change. Electric potential is potential energy per unit charge; V = U/q. For processes, ΔU = q ΔV. q = 3.0 , nC = 3.0 10 -9 , C V A = +120 , V V B = -30 , V A 3.0 nC charge moves from a point at potential +120 V to a point at potential −30 V. Find (a) the change in its potential energy, (b) work done by the electric field. medium Use ΔU = q(V B − V A ); W field = −ΔU. ΔU and W field Potential Energy of Two or More Charges For two point charges separated by distance r , potential energy is U = k , q 1 q 2 r (zero at infinity). For N charges, add over all distinct pairs: U = i<j k , q i q j r ij . The sign matters: like charges give U>0 (repulsive, unbound unless work is done to assemble); unlike charges give U<0 (attractive, bound). U = k q 1 q 2 / r Electrostatics; V( )=0 . Bring q 2 from infinity to distance r from fixed q 1 . Move quasi-statically (no kinetic energy change). U = 1 4 0 q 1 q 2 r Two-charge energy: U = k q1 q2 / r; for many charges, sum over all distinct pairs. Energy of N charges Count each unordered pair once to avoid double-counting. This equation determines the potential energy stored in a dipole when placed within a uniform external electric field. A negative total U means the configuration is bound; you must supply at least |U| energy to take all charges to infinity. If U>0 , the system already requires external work to assemble and tends to explode apart if released. Sign logic: q 1 q 2>0 gives U>0 (repulsion); q 1 q 2<0 gives U<0 (attraction). The magnitude scales inversely with separation: halving r doubles |U| . remember Three charges are at the vertices of an equilateral triangle of side 0.30 m: q1 = +2.0 µC, q2 = −3.0 µC, q3 = +4.0 µC. Find the total electrostatic potential energy of the configuration. hard U = k ( q 1 q 2 a + q 2 q 3 a + q 3 q 1 a ) q 1 = +2.0 , C = 2.0 10 -6 , C q 2 = -3.0 , C = -3.0 10 -6 , C q 3 = +4.0 , C = 4.0 10 -6 , C a = 0.30 , m k = 8.99 10 9 , N ,m 2 /C 2 Total U False. Potential can be non-zero and even large while field is zero (e.g., inside a uniformly charged spherical shell, E=0 but V= constant 0 ). If the electric field at a point is zero, the potential must be zero there. In electrostatics, the field is conservative, so work depends only on the endpoints: W = q ,[V(A)-V(B)] . Work done by the electric field depends on the path taken between two points. Dipole Potential and Energy in a Uniform Field A short dipole with moment p has potential V(r, ) = 1 4 0 , p r 2 at far points ( r dipole length). In a uniform field E , the dipole’s potential energy is U = - p E = -pE , minimum when p aligns with E . Torque is = p E , tending to align the dipole. Dipole potential (far field) Minimum at = 0 (aligned), maximum at = (anti-aligned). Dipole energy in uniform field Conductors, Spherical Shells, and Potential Maps This formula applies to the energy stored in an ideal capacitor when a potential difference (voltage) is applied across its plates. In electrostatic equilibrium, the electric field inside a conductor is zero, so the potential is constant throughout its volume and across its surface. Any excess charge resides on the outer surface. For a uniformly charged thin spherical shell of radius R , V(r) outside is that of a point charge at the center, and inside it is constant: V(r<R) = kQ/R . Potential of a thin spherical shell Field is zero inside; potential is flat and equals its surface value. From r=0 to r=R, V is constant at kQ/R; for r>R, V falls as kQ/r toward 0. custom control control V(r) derived Potential of a charged thin spherical shell: flat inside, 1/r outside. r=0 to R V=kQ/R Flat region inside Continuous at surface r=R V=kQ/R V→0 r→∞ Reference V0 to lower Potential V Radial distance r Inside any conductor at electrostatic equilibrium: E=0 and V= constant. Thus, moving a charge anywhere inside costs zero work (no change in potential). tip Field is force per unit charge, E = F/q. Combined with V, use E = −∇V to connect force, field, and potential. General field–potential integral Quantity Nature Adds how? Unit Key formula Typical use Electric Field (E) Vector Vector sum N/C E = −∇V; for point charge, E = k|q|/ r 2 Force direction, acceleration Electric Potential (V) Scalar Algebraic sum V (J/C) V = U/q; for point charge, V = kq/r Work, energy changes Potential Energy (U) Scalar Pairwise sum for systems U = qV; for two charges, U = k q1 q2/r Stability, binding energy Downhill Drives Direction: E points from higher V to lower V, perpendicular to equipotentials. Up the hill (against E) needs external work; downhill (with E) releases energy. q 1 = +6.0 , nC at x=-0.10 , m q 2 = -6.0 , nC at x=+0.10 , m V(x) = k ( q 1 |x+0.10| + q 2 |x-0.10| ) Set k(q1/|x+0.10| + q2/|x-0.10|) = 0 ⇒ q1/|x+0.10| = − q2/|x-0.10|. medium Two charges +6.0 nC at x = −0.10 m and −6.0 nC at x = +0.10 m lie on the x-axis. Find the point(s) on the x-axis (x ≠ ±0.10 m) where the potential is zero. x where V=0 Notice the contrast: at the midpoint of an equal-and-opposite pair, V=0 but E 0 (fields add, pointing from + to −). This is a classic conceptual pitfall. In V = k q / r and U = k q1 q2 / r, the denominator should be r 2 like the force. No. Potential and potential energy come from integrating the inverse-square force; the integral produces a 1/r dependence, not 1/ r 2 . neet-alert Reference level matters. If a conductor is grounded (V = 0), recompute potentials and energies relative to that condition; do not blindly use V(∞)=0. Use V = k q / r with correct sign; superpose for many charges. Track energy changes with ΔU = q ΔV for moves between equipotentials. Configuration energy equals the sum over all charge pairs. Worked strategy checklist: (1) Choose or confirm the reference for potential (usually infinity or ground). (2) Compute potentials by superposition; keep signs. (3) Convert to energy via U = qV for a test charge or U= k q i q j/r ij for a configuration. (4) Use U = q , V for work and energy changes. (5) For direction questions, recover field by E = - V . (a) E, (b) ΔU, (c) W field Plate separation d = 5.0 , mm = 5.0 10 -3 , m V L = +200 , V , V R = 0 , V q = +1.0 , nC = 1.0 10 -9 , C Two large parallel plates 5.0 mm apart are maintained at potentials +200 V (left plate) and 0 V (right plate). A +1.0 nC charge is moved from a point 1.0 mm from the left plate to a point 1.0 mm from the right plate. Find (a) the uniform electric field between the plates (magnitude and direction), (b) the change in potential energy of the charge, and (c) the work done by the field. mixed hard Uniform field: E = −ΔV/d (direction from high to low potential, left to right). Use ΔU = q ΔV for the move. neet-alert Distance in the nth second–style trap, electrical edition: When using V = kq/r, r is the straight-line distance from the charge to the point, not along any path. Do not square r in potential or energy formulas. Edge cases and limits: (1) As r 0 for a point charge, |V| and |U| (idealization; real charges have finite size). (2) As r , V 0 and pairwise U 0 . (3) If total charge is inside a conducting enclosure (Faraday cage), E=0 inside and V= constant; external fields do not alter the interior potential. Problem-solving scaffold: (a) Draw a clean diagram with distances; (b) Write V= k q i/r i ; (c) Evaluate numerically with units; (d) Convert to energy if needed; (e) Interpret sign and magnitude physically. Key Terms Recap Electric Potential (V) Voltage Scalar potential Energy per unit charge at a point: V=U/q . Energy of a charge or system due to configuration: U=qV or pairwise sum. Electrostatic energy Potential Energy (U) Set of points where potential is same; E surface. Equipotential Superposition (V) Potentials add algebraically: V= k q i/r i . Spatial derivative; E = - V links field and potential. Gradient Bound State Configuration with U<0 ; needs energy to separate to infinity.