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Electrostatics question

2019 · 10 Apr · Shift 1 · Q51
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Electrostatics question

2019 · 10 Apr · Shift 1 · Q51

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A uniformly charged ring of radius 3a and total charge q is placed in xy-plane centred at origin. A point charge q is moving towards the ring along the z-axis and has speed u at z = 4a. The minimum value of u such that it crosses the origin is :
  1. A
    2m(215q24πε0a)1/2\sqrt {{2 \over m}} {\left( {{2 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}m2​​(152​4πε0​aq2​)1/2
  2. B
    2m(115q24πε0a)1/2\sqrt {{2 \over m}} {\left( {{1 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}m2​​(151​4πε0​aq2​)1/2
  3. C
    2m(15q24πε0a)1/2\sqrt {{2 \over m}} {\left( {{1 \over {5}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}m2​​(51​4πε0​aq2​)1/2
  4. D
    2m(415q24πε0a)1/2\sqrt {{2 \over m}} {\left( {{4 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}m2​​(154​4πε0​aq2​)1/2
View written solutionFree

Correct answer: A

  1. Potential on the axis of a uniformly charged ring

For a ring of radius R=3aR=3aR=3a and charge qqq, the electric potential at a point on its axis at distance zzz is

V(z)=14πε0qR2+z2V(z)=\frac{1}{4\pi\varepsilon_0}\frac{q}{\sqrt{R^2+z^2}}V(z)=4πε0​1​R2+z2​q​

So here,

V(z)=14πε0q(3a)2+z2V(z)=\frac{1}{4\pi\varepsilon_0}\frac{q}{\sqrt{(3a)^2+z^2}}V(z)=4πε0​1​(3a)2+z2​q​

The moving particle also has charge qqq, so its potential energy is

U(z)=qV(z)=14πε0q29a2+z2U(z)=qV(z)=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{\sqrt{9a^2+z^2}}U(z)=qV(z)=4πε0​1​9a2+z2​q2​

  1. Initial and maximum potential energy

The particle starts at z=4az=4az=4a with speed uuu.

At z=4az=4az=4a,

=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{5a}$$ To cross the origin, it must be able to reach $z=0$. Since for like charges the potential energy increases as it approaches the center, the **maximum** potential energy on the path from $z=4a$ to $z=0$ occurs at $z=0$. Thus, $$U_0=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{3a}$$ --- 3. **Condition for minimum speed to just cross the origin** For the minimum speed, the particle just reaches the origin with zero speed. Using conservation of energy, $$\frac{1}{2}mu^2 + U_i = U_0$$ So, $$\frac{1}{2}mu^2 = U_0-U_i$$ Substitute the values: $$\frac{1}{2}mu^2 = \frac{1}{4\pi\varepsilon_0}q^2\left(\frac{1}{3a}-\frac{1}{5a}\right)$$ $$\frac{1}{2}mu^2 = \frac{1}{4\pi\varepsilon_0}q^2\left(\frac{5-3}{15a}\right)$$ $$\frac{1}{2}mu^2 = \frac{1}{4\pi\varepsilon_0}\frac{2q^2}{15a}$$ Hence, $$u=\sqrt{\frac{2}{m}\left(\frac{2}{15}\frac{q^2}{4\pi\varepsilon_0 a}\right)}$$ --- 4. **Matching with options** This matches **Option A**. $$\boxed{u=\sqrt{\frac{2}{m}}\left(\frac{2}{15}\frac{q^2}{4\pi\varepsilon_0 a}\right)^{1/2}}$$
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