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Electrostatics question

2006 · Shift 0 · Q116
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Electrostatics question

2006 · Shift 0 · Q116

JEE MainPhysicsElectrostaticsMCQ+4 / −1
Two insulating plates are both uniformly charged in such a way that the potential difference between them is V2−V1=20 V.{V_2} - {V_1} = 20\,V.V2​−V1​=20V.(i.e., plate 222 is at a higher potential). The plates are separated by d=0.1md=0.1md=0.1m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1.1.1. What is its speed when it hits plate 222? (e=1.6×10−19 C,  me=9.11×10−31 kg)\left( {e = 1.6 \times {{10}^{ - 19}}\,C,\,\,{m_e} = 9.11 \times {{10}^{ - 31}}\,kg} \right)(e=1.6×10−19C,me​=9.11×10−31kg) AIEEE 2006 Physics - Electrostatics Question 237 English
  1. A
    2.65×106 m/s2.65 \times {10^6}\,m/s2.65×106m/s
  2. B
    7.02×1012 m/s7.02 \times {10^{12}}\,m/s7.02×1012m/s
  3. C
    1.87×106 m/s1.87 \times {10^6}\,m/s1.87×106m/s
  4. D
    32×10−19 m/s32 \times {10^{ - 19}}\,m/s32×10−19m/s
View written solutionFree

Correct answer: A

  1. Use energy conservation

An electron starts from rest, so its initial kinetic energy is zero.

When it moves through a potential difference, the change in electric potential energy is converted into kinetic energy:

Kf−Ki=−ΔU=−q (V2−V1)K_f - K_i = -\Delta U = -q\,(V_2 - V_1)Kf​−Ki​=−ΔU=−q(V2​−V1​)

For an electron,

q=−eq = -eq=−e

So,

Kf=−(−e)(V2−V1)=e(V2−V1)K_f = -(-e)(V_2 - V_1) = e(V_2 - V_1)Kf​=−(−e)(V2​−V1​)=e(V2​−V1​)

Given:

V2−V1=20 VV_2 - V_1 = 20\,\text{V}V2​−V1​=20V

Hence,

Kf=e⋅20=(1.6×10−19)(20)K_f = e \cdot 20 = (1.6\times 10^{-19})(20)Kf​=e⋅20=(1.6×10−19)(20)

Kf=3.2×10−18 JK_f = 3.2\times 10^{-18}\,\text{J}Kf​=3.2×10−18J

  1. Relate kinetic energy to speed

Kf=12mev2K_f = \frac{1}{2}m_ev^2Kf​=21​me​v2

Therefore,

12mev2=3.2×10−18\frac{1}{2}m_ev^2 = 3.2\times 10^{-18}21​me​v2=3.2×10−18

v2=2×3.2×10−189.11×10−31v^2 = \frac{2\times 3.2\times 10^{-18}}{9.11\times 10^{-31}}v2=9.11×10−312×3.2×10−18​

v2=6.4×10−189.11×10−31v^2 = \frac{6.4\times 10^{-18}}{9.11\times 10^{-31}}v2=9.11×10−316.4×10−18​

v2≈7.02×1012v^2 \approx 7.02\times 10^{12}v2≈7.02×1012

Now take square root:

v=7.02×1012v = \sqrt{7.02\times 10^{12}}v=7.02×1012​

v≈2.65×106 m/sv \approx 2.65\times 10^6\,\text{m/s}v≈2.65×106m/s

  1. Check options
  • A: 2.65×106 m/s2.65 \times 10^6\,\text{m/s}2.65×106m/s ✅
  • B: 7.02×1012 m/s7.02 \times 10^{12}\,\text{m/s}7.02×1012m/s is actually v2v^2v2, not vvv
  • C: incorrect
  • D: incorrect

Therefore, the correct option is:

A\boxed{\text{A}}A​

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