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Capacitor question

2025 · 22 Jan · Shift 2 · Q54

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Capacitor
2025 · 22 Jan · Shift 2 · Q54

Capacitor question

2025 · 22 Jan · Shift 2 · Q54

JEE MainPhysicsCapacitorMCQ+4 / −1

Which one of the following is the correct dimensional formula for the capacitance in F ?

\mathrm{M}, \mathrm{L}, \mathrm{T}

and

C

stand for unit of mass, length, time and charge,

A
$[\mathrm{F}]=\left[\mathrm{CM}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2\right]$
B
$[\mathrm{F}]=\left[\mathrm{C}^2 \mathrm{M}^{-2} \mathrm{~L}^2 \mathrm{~T}^2\right]$
C
$[\mathrm{F}]=\left[\mathrm{C}^2 \mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2\right]$
D
$[\mathrm{F}]=\left[\mathrm{CM}^{-2} \mathrm{~L}^{-2} \mathrm{~T}^{-2}\right]$

View written solutionFree

Correct answer: C

Use the definition of capacitance

Capacitance is defined as

C_{\text{ap}}=\frac{Q}{V}

where $Q$ is charge and $V$ is potential difference.

So, dimensionally,

[\text{capacitance}] = \frac{[Q]}{[V]}

Find the dimensional formula of potential

Potential difference is

V = \frac{\text{work}}{\text{charge}}

Now,

[\text{work}] = [\text{force}][\text{distance}] = [MLT^{-2}][L] = [ML^2T^{-2}]

Hence,

[V] = \frac{[ML^2T^{-2}]}{[C]}

Now find capacitance dimension

[F] = \frac{[C]}{[V]} = \frac{[C]}{[ML^2T^{-2}/C]}

[F] = [C]\cdot \frac{[C]}{[ML^2T^{-2}]}

[F] = [C^2M^{-1}L^{-2}T^2]

Compare with options

A: $[CM^{-1}L^{-2}T^2]$ ❌
B: $[C^2M^{-2}L^2T^2]$ ❌
C: $[C^2M^{-1}L^{-2}T^2]$ ✅
D: $[CM^{-2}L^{-2}T^{-2}]$ ❌

Therefore, the correct option is:

\boxed{\text{C}}

Comparison with stored answer

Stored correct answer: C

My derived answer: C

So, the answers agree.

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