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Electrostatics question

2021 · 24 Feb · Shift 2 · Q54
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  5. /2021 · 24 Feb · Shift 2 · Q54

Electrostatics question

2021 · 24 Feb · Shift 2 · Q54

JEE MainPhysicsElectrostaticsMCQ+4 / −1
Two electrons each are fixed at a distance '2d'. A third charge proton placed at the midpoint is displaced slightly by a distance x (x << d) perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency : (m = mass of charged particle)
  1. A
    (2q2πε0md3)12{\left( {{{2{q^2}} \over {\pi {\varepsilon _0}m{d^3}}}} \right)^{{1 \over 2}}}(πε0​md32q2​)21​
  2. B
    (q22πε0md3)12{\left( {{{{q^2}} \over {2\pi {\varepsilon _0}m{d^3}}}} \right)^{{1 \over 2}}}(2πε0​md3q2​)21​
  3. C
    (2πε0md3q2)12{\left( {{{2\pi {\varepsilon _0}m{d^3}} \over {{q^2}}}} \right)^{{1 \over 2}}}(q22πε0​md3​)21​
  4. D
    (πε0md32q2)12{\left( {{{\pi {\varepsilon _0}m{d^3}} \over {2{q^2}}}} \right)^{{1 \over 2}}}(2q2πε0​md3​)21​
View written solutionFree

Correct answer: B

  1. Set up the configuration

Two electrons are fixed symmetrically, and the distance between them is 2d2d2d. Hence we may place them at

(−d,0)and(d,0)(-d,0) \quad \text{and} \quad (d,0)(−d,0)and(d,0)

A proton is initially at the midpoint (0,0)(0,0)(0,0) and is displaced slightly perpendicular to the line joining the electrons, say to

(0,x),x≪d(0,x), \qquad x \ll d(0,x),x≪d

We need the restoring force on the proton.

  1. Force due to one electron on the proton

Distance from the proton to either electron is

r=d2+x2r = \sqrt{d^2 + x^2}r=d2+x2​

Magnitude of force due to one electron:

F1=14πε0q2r2F_1 = \frac{1}{4\pi \varepsilon_0} \frac{q^2}{r^2}F1​=4πε0​1​r2q2​

This force is attractive, directed toward the electron.

Its vertical component is

= \frac{1}{4\pi \varepsilon_0} \frac{q^2 x}{r^3}$$ Since the force is toward the electron, the vertical component is downward, i.e. negative: $$F_{1y} = -\frac{1}{4\pi \varepsilon_0} \frac{q^2 x}{(d^2+x^2)^{3/2}}$$ --- 3. **Net force due to both electrons** The horizontal components cancel by symmetry. The vertical components add: $$F_y = 2F_{1y} = -\frac{2}{4\pi \varepsilon_0} \frac{q^2 x}{(d^2+x^2)^{3/2}}$$ So, $$F_y = -\frac{1}{2\pi \varepsilon_0} \frac{q^2 x}{(d^2+x^2)^{3/2}}$$ --- 4. **Use the small displacement approximation** Since $x \ll d$, $$ (d^2+x^2)^{3/2} \approx d^3 $$ Therefore, $$F_y \approx -\frac{q^2}{2\pi \varepsilon_0 d^3}x$$ This is of the form $$F = -kx$$ with effective spring constant $$k = \frac{q^2}{2\pi \varepsilon_0 d^3}$$ --- 5. **Angular frequency of SHM** For simple harmonic motion, $$\omega = \sqrt{\frac{k}{m}}$$ Hence, $$\omega = \left(\frac{q^2}{2\pi \varepsilon_0 m d^3}\right)^{1/2}$$ --- 6. **Match with options** This matches **Option B**: $$\boxed{\left(\frac{q^2}{2\pi \varepsilon_0 m d^3}\right)^{1/2}}$$
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