Question

Find the dimensional formulae of
(a) the charge Q,
(b) the potential V,
(c) the capacitance C, and
(d) the resistance R.

Concept of Physics - 1 , HC VERMA , Chapter "Introduction to Physics"

Answer 1

a) Q=IT
(Q)=M^0L^0A^1T^1
Where A is amphere

b) W=VIT
(V)=(M^1L^2T^-2)/(AT^1)=(M^1L^2T^-3A^-1) is answer

c) Q=CV
C=Q/V=IT/W/IT=I²T²/W
C=A²T²/ML^2T^-2=(M^-1L^-2T^4A^2) is answer

d) V= IR
R= V/I =W/IT/I= W/I^2T
R =(M^1L^2T^-2)/(A^2T^1)
R=( M^1L^2T^-3A^-2) is answer

Answer 2

Hello Dear.

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(a)
∵ Current(I) = Charge (Q) ÷ Time(t)
∴ Q = I × t

For Finding the Dimension of the Q,
We know, Dimension of I = I
& Dimension of Time = T

∴ [Q] = IT

Hence, the Dimension of the Charge Q [Q] = IT.

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(b) 
∵ Potential (V) = Work(W) ÷ Charge(Q)
∴ V = W/It

Now, Let us finding the Dimension,

[V] = (ML²T⁻²) ÷ (IT)
[V] = ML² I⁻¹ T⁻³

Hence, the dimension of Potential is ML² I⁻¹ T⁻³.

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(c) ∵ Q = C × V
Where, Q = Charge, C = Capitance and V = Potential.

∴ C = Q/V

Now, We know,
 the Dimension of Q = IT    [Calculated Above]
& Dimension of V = ML² I⁻¹ T⁻³.  [Calculated Above]

∴ [C] = (IT) ÷ (ML² I⁻¹ T⁻³)
⇒ [C] = M⁻¹ L⁻² I² T⁴


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(d) From the Ohm's law,

 V = I × R
∴ R = V/I

Dimension of V = M L² I⁻¹ T⁻³  [Calculated Above]
& Dimension of I = I

∴ [R] = (M L² I⁻¹ T⁻³) ÷ I
⇒ [R] = ML² I⁻² T⁻³

Hence, the Dimension of the Resistance is ML² I⁻² T⁻³.


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Hope it helps.