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PhysicsElectromagnetismElectrostaticsAdvanced Applications: Coulomb's Law & Simple Harmonic Motion

Advanced Applications: Coulomb's Law & Simple Harmonic Motion

Bridging electrostatics and kinematics: Analyzing stable and unstable equilibrium for point charges.

Advanced Applications: Coulomb's Law & Simple Harmonic Motion

While the fundamental statement of Coulomb's Law dictates the force between two stationary charges, JEE Advanced problems rarely stop there. A common archetype involves placing a system of charges in equilibrium and analyzing what happens when that equilibrium is slightly disturbed.

The Two-Charge System: Axial and Equatorial Displacement

Consider two identical positive point charges, +Q+Q, fixed at positions (d,0)(d, 0) and (d,0)(-d, 0). A third test charge, qq, is placed at the origin (0,0)(0,0). By symmetry, the net electrostatic force on qq is zero. The system is in equilibrium.

However, the nature of this equilibrium depends entirely on the sign of qq and the axis of displacement.

Case 1: Displacing a Positive Charge +q+q along the X-axis (Axial)

Let's displace +q+q by a small distance xx (xdx \ll d) along the positive x-axis. The force from the right charge increases (distance decreases to dxd-x), and the force from the left charge decreases (distance increases to d+xd+x).

The net restoring force Fnet\vec{F}_{net} directed towards the origin is:

Fnet=14πϵ0Qq(dx)214πϵ0Qq(d+x)2F_{net} = \frac{1}{4\pi\epsilon_0} \frac{Qq}{(d-x)^2} - \frac{1}{4\pi\epsilon_0} \frac{Qq}{(d+x)^2}

Factoring out the constants and d2d^2:

Fnet=Qq4πϵ0d2[(1xd)2(1+xd)2]F_{net} = \frac{Qq}{4\pi\epsilon_0 d^2} \left[ \left(1 - \frac{x}{d}\right)^{-2} - \left(1 + \frac{x}{d}\right)^{-2} \right]

Since xdx \ll d, we use the binomial approximation (1+z)n1+nz(1+z)^n \approx 1 + nz:

FnetQq4πϵ0d2[(1+2xd)(12xd)]=Qq4πϵ0d2(4xd)F_{net} \approx \frac{Qq}{4\pi\epsilon_0 d^2} \left[ \left(1 + 2\frac{x}{d}\right) - \left(1 - 2\frac{x}{d}\right) \right] = \frac{Qq}{4\pi\epsilon_0 d^2} \left( \frac{4x}{d} \right)

Fnet=(Qqπϵ0d3)xF_{net} = -\left( \frac{Qq}{\pi\epsilon_0 d^3} \right) x

Because FnetxF_{net} \propto -x, the charge executes Simple Harmonic Motion (SHM). The effective spring constant is keff=Qqπϵ0d3k_{eff} = \frac{Qq}{\pi\epsilon_0 d^3}, giving a time period of:

T=2πmkeff=2πmπϵ0d3QqT = 2\pi \sqrt{\frac{m}{k_{eff}}} = 2\pi \sqrt{\frac{m \pi \epsilon_0 d^3}{Qq}}

Equilibrium Summary Matrix

Charge SignDisplacement AxisNet Force DirectionType of EquilibriumExecutes SHM?
+q+qX-axis (Axial)Towards originStableYes
+q+qY-axis (Equatorial)Away from originUnstableNo
q-qX-axis (Axial)Away from originUnstableNo
q-qY-axis (Equatorial)Towards originStableYes

JEE Pro-Tip: Earnshaw's Theorem dictates that a charged particle cannot be in a stable equilibrium in all three spatial dimensions relying solely on electrostatic forces. If it is stable along the x-axis, it must be unstable along the y or z-axis.