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Capacitor question

2021 · 27 Jul · Shift 2 · Q50
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  5. /2021 · 27 Jul · Shift 2 · Q50

Capacitor question

2021 · 27 Jul · Shift 2 · Q50

JEE MainPhysicsCapacitorMCQ+4 / −1
Match List I with List II.

List - I List - II
(a) Capacitance, C (i) M1L1T−3A−1{M^1}{L^1}{T^{ - 3}}{A^{ - 1}}M1L1T−3A−1
(b) Permittivity of free space, ε0{\varepsilon _0}ε0​ (ii) M−1L−3T4A2{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}M−1L−3T4A2
(c) Permeability of free space, μ0{\mu _0}μ0​ (iii) M−1L−2T4A2{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}M−1L−2T4A2
(d) Electric field, E (iv) M1L1T−2A−2{M^1}{L^1}{T^{ - 2}}{A^{ - 2}}M1L1T−2A−2


Choose the correct answer from the options given below
  1. A
    (a) →\to→(iii), (b) →\to→(ii), (c) →\to→(iv), (d) →\to→ (i)
  2. B
    (a) →\to→(iii), (b) →\to→(iv), (c) →\to→(ii), (d) →\to→ (i)
  3. C
    (a) →\to→(iv), (b) →\to→(ii), (c) →\to→(iii), (d) →\to→ (i)
  4. D
    (a) →\to→(iv), (b) →\to→(iii), (c) →\to→(ii), (d) →\to→ (i)
View written solutionFree

Correct answer: A

  1. Find the dimensions of each physical quantity

We use base dimensions [M,L,T,A][M, L, T, A][M,L,T,A].

  1. (a) Capacitance CCC

Capacitance is

C=QVC = \frac{Q}{V}C=VQ​

Now,

[Q]=[AT][Q] = [AT][Q]=[AT]

and

[V]=workcharge=[ML2T−2][AT]=[ML2T−3A−1][V] = \frac{\text{work}}{\text{charge}} = \frac{[ML^2T^{-2}]}{[AT]} = [ML^2T^{-3}A^{-1}][V]=chargework​=[AT][ML2T−2]​=[ML2T−3A−1]

Therefore,

[C]=[AT][ML2T−3A−1]=[M−1L−2T4A2][C] = \frac{[AT]}{[ML^2T^{-3}A^{-1}]} = [M^{-1}L^{-2}T^4A^2][C]=[ML2T−3A−1][AT]​=[M−1L−2T4A2]

So,

(a)→(iii)(a) \to (iii)(a)→(iii)
  1. (b) Permittivity of free space ε0\varepsilon_0ε0​

From Coulomb's law,

F=14πε0q1q2r2F = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}F=4πε0​1​r2q1​q2​​

So,

[ε0]=[q2][Fr2][\varepsilon_0] = \frac{[q^2]}{[Fr^2]}[ε0​]=[Fr2][q2]​

Now,

[q2]=[A2T2][q^2] = [A^2T^2][q2]=[A2T2] [Fr2]=[MLT−2]⋅[L2]=[ML3T−2][Fr^2] = [MLT^{-2}]\cdot[L^2] = [ML^3T^{-2}][Fr2]=[MLT−2]⋅[L2]=[ML3T−2]

Hence,

[ε0]=[M−1L−3T4A2][\varepsilon_0] = [M^{-1}L^{-3}T^4A^2][ε0​]=[M−1L−3T4A2]

So,

(b)→(ii)(b) \to (ii)(b)→(ii)
  1. (c) Permeability of free space μ0\mu_0μ0​

Using

μ0ε0=1c2\mu_0\varepsilon_0 = \frac{1}{c^2}μ0​ε0​=c21​

Since

[c2]=[L2T−2]⇒[1c2]=[L−2T2][c^2] = [L^2T^{-2}] \Rightarrow \left[\frac{1}{c^2}\right] = [L^{-2}T^2][c2]=[L2T−2]⇒[c21​]=[L−2T2]

Thus,

[μ0]=[L−2T2][ε0][\mu_0] = \frac{[L^{-2}T^2]}{[\varepsilon_0]}[μ0​]=[ε0​][L−2T2]​

Substitute

[ε0]=[M−1L−3T4A2][\varepsilon_0] = [M^{-1}L^{-3}T^4A^2][ε0​]=[M−1L−3T4A2]

So,

[μ0]=[L−2T2]⋅[ML3T−4A−2]=[MLT−2A−2][\mu_0] = [L^{-2}T^2] \cdot [ML^3T^{-4}A^{-2}] = [MLT^{-2}A^{-2}][μ0​]=[L−2T2]⋅[ML3T−4A−2]=[MLT−2A−2]

So,

(c)→(iv)(c) \to (iv)(c)→(iv)
  1. (d) Electric field EEE

Electric field is force per unit charge:

E=FqE = \frac{F}{q}E=qF​

Therefore,

[E]=[MLT−2][AT]=[MLT−3A−1][E] = \frac{[MLT^{-2}]}{[AT]} = [MLT^{-3}A^{-1}][E]=[AT][MLT−2]​=[MLT−3A−1]

So,

(d)→(i)(d) \to (i)(d)→(i)
  1. Final matching

Thus the correct matching is:

(a)→(iii),(b)→(ii),(c)→(iv),(d)→(i)(a) \to (iii), \quad (b) \to (ii), \quad (c) \to (iv), \quad (d) \to (i)(a)→(iii),(b)→(ii),(c)→(iv),(d)→(i)

This corresponds to Option A.

  1. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So, the derived answer agrees with the stored answer.

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