Physics Lab

Unit 10: Electrostatics and Current Electricity

This is the highest-weight unit in NEET Physics — expect 4–5 MCQs combined. Electrostatics yields questions on point-charge force, dipoles, Gauss's law applications, and capacitor networks. Current Electricity tests Ohm's law, resistor networks, Kirchhoff's laws, and Wheatstone/potentiometer.

Concept Map

Electrostatics

  • Coulomb's law, principle of superposition
  • Electric field of a point charge, dipole, ring, infinite line, plane
  • Gauss's law and applications
  • Electric potential and equipotential surfaces
  • Capacitors: parallel-plate, dielectrics, energy, combinations

Current Electricity

  • Drift velocity and Ohm's law
  • Resistivity, temperature dependence
  • Resistor combinations
  • EMF, internal resistance
  • Kirchhoff's voltage/current laws
  • Wheatstone bridge, meter bridge, potentiometer
  • Heating effect (Joule's law)

Topic 1: Electric Charge and Coulomb's Law

Sub-topic A: Properties of Charge

  • Charge is quantised: q=neq = n e with e=1.6×1019 Ce = 1.6 \times 10^{-19}\ \text{C}.
  • Charge is conserved: net charge of isolated system is constant.
  • Charge is invariant: same in all inertial frames.
  • Like charges repel, unlike attract.

Sub-topic B: Coulomb's Law

Force between two point charges q1q_1 and q2q_2 separated by rr:

F=14πε0q1q2r2,k=14πε0=9×109 N m2/C2.F = \frac{1}{4\pi\varepsilon_0}\,\frac{q_1 q_2}{r^2}, \quad k = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \text{N m}^2/\text{C}^2.

In a medium of dielectric constant KK (relative permittivity), force is reduced by factor KK.

Topic 2: Electric Field

Sub-topic A: Definition

E=F/q0,\vec E = \vec F/q_0,

with q0q_0 a small positive test charge. Unit: N/C or V/m.

Sub-topic B: Field of a Point Charge

E=kqr2,E=kqr2r^.E = \frac{kq}{r^2}, \quad \vec E = \frac{kq}{r^2}\hat r.

Sub-topic C: Field of a Dipole

Two charges ±q\pm q separated by small distance 2a2a. Dipole moment p=q2a\vec p = q \cdot 2a from q-q to +q+q.

  • Axial point (along dipole, distance rr from centre):
Eaxial=14πε02pr3.E_\text{axial} = \frac{1}{4\pi\varepsilon_0}\,\frac{2 p}{r^3}.
  • Equatorial point (perpendicular to dipole):
Eeq=14πε0pr3.E_\text{eq} = \frac{1}{4\pi\varepsilon_0}\,\frac{p}{r^3}.

Note: Eaxial=2EeqE_\text{axial} = 2 E_\text{eq}.

Sub-topic D: Field of Continuous Distributions

For an infinite line of linear charge density λ\lambda:

E=λ2πε0r.E = \frac{\lambda}{2\pi\varepsilon_0 r}.

For an infinite sheet of surface charge density σ\sigma:

E=σ2ε0(both sides).E = \frac{\sigma}{2\varepsilon_0} \quad \text{(both sides)}.

For a conductor surface with charge density σ\sigma:

Ejust outside=σ/ε0.E_\text{just outside} = \sigma/\varepsilon_0.

For a uniformly charged ring of radius RR on axis at distance xx:

Eaxis=kqx(R2+x2)3/2.E_\text{axis} = \frac{kqx}{(R^2 + x^2)^{3/2}}.

Maximum at x=R/2x = R/\sqrt{2}.

Topic 3: Gauss's Law

EdA=qenclosedε0.\oint \vec E \cdot d\vec A = \frac{q_\text{enclosed}}{\varepsilon_0}.

Sub-topic A: Applications

  • Spherical shell of charge QQ:

    • Outside (r>Rr > R): E=kQ/r2E = kQ/r^2.
    • Inside (r<Rr < R): E=0E = 0 (cavity).
  • Solid uniformly charged sphere of radius RR and total charge QQ:

    • Outside: E=kQ/r2E = kQ/r^2.
    • Inside: E=kQr/R3E = kQr/R^3 (linear in rr).
  • Infinite cylinder: similar logic, gives E=λ/(2πε0r)E = \lambda/(2\pi\varepsilon_0 r) for r>Rr > R.

Sub-topic B: Conductors in Electrostatics

  • Inside a conductor: E=0E = 0 in equilibrium.
  • Charge resides on the surface.
  • Field just outside conductor: E=σ/ε0E = \sigma/\varepsilon_0 (perpendicular).
  • Cavity inside conductor: E=0E = 0 inside cavity (Faraday cage).

Topic 4: Electric Potential

Sub-topic A: Definition

VAVB=BAEd,V=rEd.V_A - V_B = -\int_B^A \vec E \cdot d\vec\ell, \quad V = -\int_\infty^r \vec E \cdot d\vec\ell.

For a point charge: V(r)=kq/rV(r) = kq/r.

Unit: volt (V) == J/C.

Sub-topic B: Work and Energy

Work done by an external agent to bring charge qq from \infty to r\vec r: W=qV(r)W = qV(\vec r).

PE of two charges: U=kq1q2/rU = k q_1 q_2/r.

Sub-topic C: Dipole in External Field

Torque: τ=p×E\vec\tau = \vec p \times \vec E, τ=pEsinθ\vert \tau\vert = pE\sin\theta.

Potential energy: U=pE=pEcosθU = -\vec p \cdot \vec E = -pE\cos\theta.

Stable equilibrium at θ=0\theta = 0; unstable at θ=π\theta = \pi.

Sub-topic D: Equipotential Surfaces

  • Surfaces of constant VV.
  • E\vec E \perp equipotential surface.
  • No work done in moving charge along equipotential.
  • For point charge: concentric spheres.
  • For uniform field: planes perpendicular to field.

Topic 5: Capacitance

Sub-topic A: Definition

C=Q/V.C = Q/V.

Unit: farad (F).

Sub-topic B: Parallel-Plate Capacitor

For plates of area AA separated by dd in vacuum:

C0=ε0A/d.C_0 = \varepsilon_0 A/d.

With dielectric of constant KK filling the gap:

C=KC0=Kε0A/d.C = K C_0 = K\varepsilon_0 A/d.

Partial filling (thickness tt of dielectric):

C=ε0Adt+t/K.C = \frac{\varepsilon_0 A}{d - t + t/K}.

Sub-topic C: Energy Stored

U=12CV2=12Q2C=12QV.U = \tfrac{1}{2} CV^2 = \tfrac{1}{2}\,\frac{Q^2}{C} = \tfrac{1}{2}QV.

Energy per unit volume in field: u=12ε0E2u = \tfrac{1}{2}\varepsilon_0 E^2.

Sub-topic D: Combinations

CombinationFormula
Series1/Ceq=1/C1+1/C21/C_\text{eq} = 1/C_1 + 1/C_2
ParallelCeq=C1+C2C_\text{eq} = C_1 + C_2

In series, charges are equal across each capacitor. In parallel, voltages are equal.

Sub-topic E: Connection / Sharing of Charge

Two capacitors with Q1Q_1 and Q2Q_2 on C1C_1 and C2C_2 connected in parallel:

Vcommon=(Q1+Q2)/(C1+C2).V_\text{common} = (Q_1 + Q_2)/(C_1 + C_2).

Heat dissipated:

ΔU=C1C2(V1V2)22(C1+C2).\Delta U = \frac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)}.

Topic 6: Current Electricity

Sub-topic A: Current and Drift Velocity

I=nAevd,I = nAev_d,

with nn free electron density, AA cross-section, vdv_d drift speed (~ 104 m/s10^{-4}\ \text{m/s}).

Drift velocity:

vd=eEτ/m,τ=mean free time.v_d = eE\tau/m, \quad \tau = \text{mean free time}.

Sub-topic B: Ohm's Law

V=IR,R=ρL/A,ρ=m/(ne2τ).V = I R, \quad R = \rho L/A, \quad \rho = m/(n e^2 \tau).

Resistivity ρ\rho is material property; RR is geometry-dependent.

Sub-topic C: Temperature Dependence

R=R0(1+αΔT),R = R_0 (1 + \alpha\Delta T),

with α>0\alpha > 0 for metals (resistance increases with T). For semiconductors, α<0\alpha < 0 (resistance falls with T).

Sub-topic D: Resistor Combinations

CombinationFormula
SeriesReq=R1+R2+R_\text{eq} = R_1 + R_2 + \dots
Parallel1/Req=1/R1+1/R2+1/R_\text{eq} = 1/R_1 + 1/R_2 + \dots

Topic 7: EMF and Internal Resistance

Sub-topic A: Cell Model

A real cell has emf ε\varepsilon and internal resistance rr. Terminal voltage:

V=εIr(when discharging).V = \varepsilon - I r \quad (\text{when discharging}).

When charging: V=ε+IrV = \varepsilon + Ir.

Power delivered to load RR:

PR=ε2R/(R+r)2.P_R = \varepsilon^2 R/(R + r)^2.

Maximum at R=rR = r (impedance matching): Pmax=ε2/(4r)P_\text{max} = \varepsilon^2/(4r), efficiency 50%.

Sub-topic B: Cells in Series and Parallel

  • Series (same direction): εnet=ε1+ε2\varepsilon_\text{net} = \varepsilon_1 + \varepsilon_2, rnet=r1+r2r_\text{net} = r_1 + r_2.
  • Parallel (identical cells, nn of them): εnet=ε\varepsilon_\text{net} = \varepsilon, rnet=r/nr_\text{net} = r/n.

Topic 8: Kirchhoff's Laws

Sub-topic A: Junction Rule (KCL)

At any junction, Iin=Iout\sum I_\text{in} = \sum I_\text{out} (conservation of charge).

Sub-topic B: Loop Rule (KVL)

Around any closed loop, ΔV=0\sum \Delta V = 0 (conservation of energy).

Sign convention: traverse loop and add +ε+\varepsilon if from - to ++ of cell, subtract IR drops if traversing in current direction.

Topic 9: Wheatstone Bridge and Meter Bridge

Sub-topic A: Wheatstone Bridge

Four resistors P,Q,R,SP, Q, R, S in a bridge. Balance condition:

PQ=RS.\frac{P}{Q} = \frac{R}{S}.

When balanced, no current through galvanometer; bridge insensitive to galvanometer resistance and cell EMF.

Sub-topic B: Meter Bridge

A 1 m wire of uniform resistance. Bridge balance at length \ell from one end gives

RS=100.\frac{R}{S} = \frac{\ell}{100 - \ell}.

Sub-topic C: Potentiometer

A long uniform wire with constant current. Potential gradient k=V/Lk = V/L. To compare two emfs ε1\varepsilon_1 and ε2\varepsilon_2 with balance lengths 1\ell_1 and 2\ell_2:

ε1ε2=12.\frac{\varepsilon_1}{\varepsilon_2} = \frac{\ell_1}{\ell_2}.

Advantage of potentiometer over voltmeter: at balance no current is drawn from the cell — so it measures true emf, not terminal voltage.

To find internal resistance: balance with cell alone (length 1\ell_1), then with cell in parallel with external RR (length 2\ell_2):

r=R(1/21).r = R(\ell_1/\ell_2 - 1).

Topic 10: Heating Effect of Current (Joule's Law)

Power dissipated in resistor:

P=I2R=V2/R=VI.P = I^2 R = V^2/R = VI.

Energy: E=PtE = Pt (joules).

In household: 1 unit = 1 kWh = 3.6×1063.6 \times 10^6 J.

NEET Pattern MCQ Tips

  • Coulomb's law / 3-charge: forces along triangle.
  • Dipole field: axial vs equatorial (factor 2).
  • Gauss's law: choose Gaussian surface aligning with symmetry.
  • Capacitor energy: before vs after dielectric insertion.
  • Capacitor sharing: heat lost = C1C2(V1V2)2/[2(C1+C2)]C_1 C_2 (V_1 - V_2)^2/[2(C_1 + C_2)].
  • Resistor network: identify series/parallel.
  • Kirchhoff numerical: 2-loop circuit.
  • Wheatstone / Meter bridge: balance condition.
  • Potentiometer: cell emf comparison.
  • Drift velocity: I=nAevdI = nAev_d.
  • Assertion-Reason: equipotential surfaces, current direction.

Common Confusions and Traps

  • Like charges of +q+q have U=+kq2/r>0U = +kq^2/r > 0; unlike have U<0U < 0.
  • Field inside a charged spherical shell is zero but potential is non-zero (constant equal to surface value).
  • Dielectric constant KK increases capacitance (C=KC0C = KC_0).
  • When a battery is disconnected and dielectric inserted: QQ stays same, VV decreases, EE decreases, energy decreases.
  • When battery remains connected and dielectric inserted: VV stays same, QQ increases, EE stays same, energy increases.
  • The drift velocity is very small (~ 104 m/s10^{-4}\ \text{m/s}) yet current travels nearly at cc because the electric field establishes itself quickly.
  • Resistivity is a property of the material; resistance is a property of the piece.
  • A galvanometer does not read EMF directly; one needs a potentiometer.
  • Internal resistance lowers the terminal voltage below EMF when current flows.

Quick Revision Card

  • F=kq1q2/r2F = kq_1q_2/r^2; k=9×109k = 9 \times 10^9 Nm²/C².
  • Epoint=kq/r2E_\text{point} = kq/r^2; Vpoint=kq/rV_\text{point} = kq/r.
  • Dipole axial: 2kp/r32kp/r^3; equatorial: kp/r3kp/r^3.
  • Infinite line: E=λ/(2πε0r)E = \lambda/(2\pi\varepsilon_0 r).
  • Infinite sheet: E=σ/(2ε0)E = \sigma/(2\varepsilon_0).
  • Charged shell: Ein=0E_\text{in} = 0, Eout=kQ/r2E_\text{out} = kQ/r^2.
  • Cparallel-plate=ε0A/dC_\text{parallel-plate} = \varepsilon_0 A/d; with dielectric ×K\times K.
  • Series caps: 1/Ceq1/C_\text{eq}; parallel: add.
  • Energy: U=Q2/(2C)=CV2/2U = Q^2/(2C) = CV^2/2.
  • I=nAevdI = nAev_d; ρ=m/(ne2τ)\rho = m/(ne^2\tau).
  • Series R: add; parallel: 1/Req1/R_\text{eq}.
  • Wheatstone balance: P/Q=R/SP/Q = R/S.
  • Potentiometer: ε1/ε2=1/2\varepsilon_1/\varepsilon_2 = \ell_1/\ell_2.
  • P=I2R=V2/RP = I^2 R = V^2/R.

Worked NEET Examples

Example 1: Force Between Charges

Two charges +10 μ+10\ \muC and +20 μ+20\ \muC separated by 10 cm in vacuum. Force:

F=(9×109)(10×106)(20×106)/(0.1)2=(9×200×103)/(0.01)=180 N.F = (9 \times 10^9)(10 \times 10^{-6})(20 \times 10^{-6})/(0.1)^2 = (9 \times 200 \times 10^{-3})/(0.01) = 180\ \text{N}.

Example 2: Field at Centre of Square

Four equal charges +q at corners of a square of side aa. Net field at centre = 0 (by symmetry).

If the four charges are +q, +q, +q, -q, then by symmetry only the diagonals matter. Net field has component from the imbalance, pointing toward -q.

Example 3: Capacitor with Dielectric

A parallel plate capacitor (1 μF in vacuum) is connected to a 100 V battery. Dielectric of K=4K = 4 is inserted while battery is connected. New charge:

Before: Q0=CV=106×100=104Q_0 = CV = 10^{-6} \times 100 = 10^{-4} C. After: C=4μC' = 4 \muF. Q=CV=4×104Q' = C'V = 4 \times 10^{-4} C. So Q quadruples; energy quadruples; field unchanged.

If battery is disconnected first, then dielectric inserted: QQ unchanged, CC → 4C, VV → V/4, EE → E/4, energy → energy/4.

Example 4: Drift Velocity

Copper wire 1 mm² cross-section carrying 1 A. Free electron density n=8.5×1028/n = 8.5 \times 10^{28}/m³.

vd=I/(nAe)=1/(8.5×1028×106×1.6×1019)=1/(1.36×104)7.3×105 m/s.v_d = I/(nAe) = 1/(8.5 \times 10^{28} \times 10^{-6} \times 1.6 \times 10^{-19}) = 1/(1.36 \times 10^4) \approx 7.3 \times 10^{-5}\ \text{m/s}.

Very small drift velocity — yet current flows nearly instantly because the electric field is established at the speed of light.

Example 5: Wheatstone Bridge

In a Wheatstone bridge: P=4Ω,Q=8Ω,R=6ΩP = 4\Omega, Q = 8\Omega, R = 6\Omega. For balance, P/Q=R/SP/Q = R/S, so S=RQ/P=6×8/4=12 ΩS = R Q/P = 6 \times 8/4 = 12\ \Omega.

Derivations

Field of an Infinite Line of Charge (Gauss)

Cylinder of radius rr, length LL coaxial with wire. By symmetry, E is radial.

EdA=E2πrL=λL/ε0.\oint E\,dA = E \cdot 2\pi r L = \lambda L/\varepsilon_0.

So E=λ/(2πε0r)E = \lambda/(2\pi\varepsilon_0 r).

Field Near Infinite Sheet (Gauss)

Pillbox crossing the sheet, area AA each face. Flux: 2EA=σA/ε02 E A = \sigma A/\varepsilon_0, so E=σ/(2ε0)E = \sigma/(2\varepsilon_0) — independent of distance.

Field Inside a Solid Charged Sphere

For uniform volume density ρ\rho in a sphere of radius RR. At radius r<Rr < R, enclosed charge: qen=ρ(4/3)πr3q_\text{en} = \rho \cdot (4/3)\pi r^3.

E4πr2=qen/ε0=(ρ/ε0)(4/3)πr3.E \cdot 4\pi r^2 = q_\text{en}/\varepsilon_0 = (\rho/\varepsilon_0)(4/3)\pi r^3.

E=ρr/(3ε0)E = \rho r/(3\varepsilon_0). Equivalently E=kQr/R3E = kQ r/R^3.

Capacitance of Parallel Plates

For two oppositely-charged plates with σ=Q/A\sigma = Q/A:

EE between plates = σ/ε0=Q/(ε0A)\sigma/\varepsilon_0 = Q/(\varepsilon_0 A).

Voltage: V=Ed=Qd/(ε0A)V = E d = Qd/(\varepsilon_0 A).

Capacitance: C=Q/V=ε0A/dC = Q/V = \varepsilon_0 A/d.

Energy in a Capacitor

Work done to charge from 0 to QQ: W=0QVdq=0Q(q/C)dq=Q2/(2C)W = \int_0^Q V\,dq = \int_0^Q (q/C)\,dq = Q^2/(2C).

Equivalently: U=(1/2)CV2=(1/2)QVU = (1/2)CV^2 = (1/2)QV.

Drift Velocity from Ohm's Law

In a wire with electric field EE, electrons accelerate, then collide. Between collisions, gained velocity Δv=eEτ/m\Delta v = eE\tau/m (where τ\tau is mean free time). On average, vd=eEτ/mv_d = eE\tau/m.

Current density: j=nevd=ne2τE/mj = ne v_d = ne^2\tau E/m. So conductivity σ=ne2τ/m\sigma = ne^2\tau/m and resistivity ρ=m/(ne2τ)\rho = m/(ne^2\tau).

Special Topics

Field of a Uniformly Charged Ring

On the axis at distance xx: E=kQx/(R2+x2)3/2E = kQx/(R^2 + x^2)^{3/2}. Maximum at x=R/2x = R/\sqrt 2, value: kQ/(33R2/2)kQ/(3\sqrt 3 R^2/2).

Field of a Uniformly Charged Disc

On the axis at distance xx (disc radius RR):

E=(σ/2ε0)[1x/x2+R2].E = (\sigma/2\varepsilon_0)[1 - x/\sqrt{x^2 + R^2}].

Limits: x0x \to 0E=σ/(2ε0)E = \sigma/(2\varepsilon_0) (infinite sheet result). xRx \gg REkQ/x2E \to kQ/x^2 (point charge).

Energy Density of Electric Field

Energy stored per unit volume in an electric field:

u=(1/2)ε0E2.u = (1/2)\varepsilon_0 E^2.

For a charged capacitor, integrating uu over the volume between plates recovers (1/2)CV2(1/2)CV^2.

Kirchhoff Solving Strategy

Label currents in each branch. Apply junction rule (KCL) at each node. Apply loop rule (KVL) for independent loops. Solve the linear system.

For a 2-loop circuit with two unknowns, you get two equations.

Maximum Power Transfer in Battery

Power delivered to external resistance RR by cell with EMF ε\varepsilon, internal rr:

P=I2R=[ε/(R+r)]2R.P = I^2 R = [\varepsilon/(R+r)]^2 R.

dP/dR=0dP/dR = 0 gives R=rR = r, Pmax=ε2/(4r)P_\text{max} = \varepsilon^2/(4r), efficiency 50%.

Cells in Combinations

N Identical Cells in Series

εnet=Nε\varepsilon_\text{net} = N\varepsilon, rnet=Nrr_\text{net} = Nr. Current through external RR: I=Nε/(R+Nr)I = N\varepsilon/(R + Nr).

For RrR \gg r: INε/RI \approx N\varepsilon/R (use series).

N Identical Cells in Parallel

εnet=ε\varepsilon_\text{net} = \varepsilon, rnet=r/Nr_\text{net} = r/N. Current: I=ε/(R+r/N)=Nε/(NR+r)I = \varepsilon/(R + r/N) = N\varepsilon/(NR + r).

For RrR \ll r: INε/rI \approx N\varepsilon/r (use parallel).

N Cells in M Rows

A 2D array of NN cells (each in series within a row, MM rows in parallel): condition for max current is NRexternal=MrNR_\text{external} = Mr.

Heating and Power Calculations

SetupPower
Single bulb RR across VVV2/RV^2/R
Two bulbs in seriesV2/(R1+R2)V^2/(R_1 + R_2), share inversely with R
Two bulbs in paralleleach takes V2/RV^2/R separately
Heater of WW at VV used at VV'W=W(V/V)2W' = W (V'/V)^2

A common NEET question: a 100 W, 220 V bulb run on 110 V dissipates 100(110/220)2=25100 (110/220)^2 = 25 W.

Formula Sheet

QuantityFormula
Coulomb forceF=kq1q2/r2F = kq_1q_2/r^2
Field of point chargeE=kq/r2E = kq/r^2
Field of dipole (axial)E=2kp/r3E = 2kp/r^3
Field of dipole (eq.)E=kp/r3E = kp/r^3
Field of infinite lineE=λ/(2πε0r)E = \lambda/(2\pi\varepsilon_0 r)
Field of infinite sheetE=σ/(2ε0)E = \sigma/(2\varepsilon_0)
Field at conductorE=σ/ε0E = \sigma/\varepsilon_0
Field on ring axisE=kQx/(R2+x2)3/2E = kQx/(R^2+x^2)^{3/2}
Potential of point chargeV=kq/rV = kq/r
PE of two chargesU=kq1q2/rU = kq_1q_2/r
Dipole torqueτ=pEsinθ\tau = pE\sin\theta
Dipole PEU=pEcosθU = -pE\cos\theta
Gauss's lawEdA=qen/ε0\oint E\,dA = q_\text{en}/\varepsilon_0
Parallel-plate CCC=ε0A/dC = \varepsilon_0 A/d
With dielectricC=Kε0A/dC = K\varepsilon_0 A/d
Capacitor energyU=12CV2=Q2/(2C)U = \tfrac{1}{2}CV^2 = Q^2/(2C)
Series caps1/Ceq=1/Ci1/C_\text{eq} = \sum 1/C_i
Parallel capsCeq=CiC_\text{eq} = \sum C_i
CurrentI=nAevdI = nAev_d
Ohm's lawV=IRV = IR
ResistanceR=ρL/AR = \rho L/A
Series RReq=RiR_\text{eq} = \sum R_i
Parallel R1/Req=1/Ri1/R_\text{eq} = \sum 1/R_i
Terminal voltageV=εIrV = \varepsilon - Ir
Max powerPmax=ε2/(4r)P_\text{max} = \varepsilon^2/(4r) at R=rR = r
Wheatstone balanceP/Q=R/SP/Q = R/S
Meter bridgeR/S=/(100)R/S = \ell/(100-\ell)
Potentiometerε1/ε2=1/2\varepsilon_1/\varepsilon_2 = \ell_1/\ell_2
Joule heatingP=I2R=V2/RP = I^2R = V^2/R

Sub-topics

6 pages

Practice quiz

Quiz
NEET Unit 10: Electrostatics and Current Electricity — Quiz
15 questions · pick the best answer
Q1

The force between two point charges is F. If the distance between them is doubled and each charge is also doubled, the new force is:

Q2

Electric field on the axis of an electric dipole at distance r (>> dipole size) is:

Q3

Capacitance of a parallel-plate capacitor with plates of area A and separation d in vacuum is:

Q4

Three capacitors each of C are connected in parallel, and the combination is connected in series with another capacitor C. Effective capacitance:

Q5

Electric field inside a uniformly charged spherical shell is:

Q6

Energy stored in a 10 μF capacitor charged to 100 V:

Q7

Two resistors 6Ω and 3Ω in parallel give equivalent resistance:

Q8

A cell of emf 6 V and internal resistance 1 Ω is connected to an external resistance of 5 Ω. Terminal voltage:

Q9

Wheatstone bridge is balanced when:

Q10

Drift velocity of electrons in a wire is approximately:

Q11

Assertion: Resistance of a metallic conductor increases with temperature. Reason: At higher temperature, more electron-lattice collisions occur, reducing relaxation time τ.

Q12

If a dielectric (K = 4) is inserted in a parallel-plate capacitor with battery disconnected, the energy stored:

Q13

Two cells of emf 2 V each and internal resistance 1 Ω are connected in parallel to drive a current through an external resistance of 0.5 Ω. The current is:

Q14

In a potentiometer experiment, cell of EMF ε₁ balances at 75 cm. Another cell ε₂ balances at 50 cm. Then ε₁/ε₂:

Q15

Assertion: Electric field inside a conductor is zero in electrostatic equilibrium. Reason: Free charges in a conductor redistribute to cancel any internal field.