E.m.f has the same unit as electric potential, hence they are dimensionally equivalent. From Electrostatics, potential difference = kq/r where k=constant (Nm2C-2), q=charge (C), r=distance (m). Find the dimensions of P.d.

Physics
E.m.f has the same unit as electric potential, hence they are dimensionally equivalent. From Electrostatics, potential difference = kq/r where k=constant (Nm2C-2), q=charge (C), r=distance (m). Find the dimensions of P.d.

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Answer

[Pd]=ML2T−3I−1\mathrm{[Pd]} = M L^{2 T^{-3} I^{-1}}

Step 1: Potential difference from electrostatics

Pd=kqrPd = \frac{k q}{r}

where [k]=N⋅m2⋅C−2[k] = N \cdot m^2 \cdot C^{-2}, [q]=C[q] = \mathrm{C}, [r]=m[r] = \mathrm{m}.

Substitute dimensions:

[Pd]=[N⋅m2⋅C−2]⋅[C][m]=N⋅m⋅C−1.[Pd] = \frac{[N \cdot m^2 \cdot C^{-2}] \cdot [C]}{[m]} = N \cdot m \cdot C^{-1}.

Step 2: Dimensions of force

[N]=[F]=MLT−2.[N] = [F] = M L T^{-2}.

Thus,

[Pd]=(MLT−2)⋅L⋅C−1=ML2T−2C−1.[Pd] = (M L T^{-2}) \cdot L \cdot C^{-1} = M L^2 T^{-2 C^{-1}}.

Step 3: Dimensions of charge

[C]=[A⋅s]=IT.[C] = [A \cdot s] = I T.

Substitute:

[Pd]=ML2T−2⋅(IT)−1=ML2T−2⋅I−1T−1=ML2T−3I−1.[Pd] = M L^2 T^{-2} \cdot (I T)^{-1} = M L^2 T^{-2 \cdot I^{-1} T^{-1}} = M L^2 T^{-3 I^{-1}}.

Step 4: Confirm from work per unit charge
Work done =F⋅r= \mathrm{F \cdot r}, so

[work]=(MLT−2)⋅L=ML2T−2.[work] = (M L T^{-2}) \cdot L = M L^2 T^{-2}. [Pd]=[work][q]=ML2T−2IT=ML2T−3I−1.[Pd] = \frac{[work]}{[q]} = \frac{M L^2 T^{-2}}{I T} = M L^2 T^{-3 I^{-1}}.

Step 5: E.M.F. has same dimensions as electric potential difference.

[Pd]=ML2T−3I−1\boxed{[Pd] = M L^{2 T^{-3} I^{-1}}}

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Potential difference from electrostatics Pd = (k q)/(r) where [k] = N · m^2 · C^-2, [q] = C, [r] = m.

E.m.f has the same unit as electric potential, hence they are dimensionally equivalent. From Electrostatics, potential difference = kq/r where k=constant (Nm2C-2), q=charge (C), r=distance (m). Find the dimensions of P.d.
Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

ScanSolve AI Answer
Step 1: Potential difference from electrostatics Pd = (k q)/(r) where [k] = N · m^2 · C^-2, [q] = C, [r] = m. Substitute dimensions: [Pd] = [N · m^2 · C^-2] · [C][m] = N · m · C^-1. Step 2: Dimensions of force [N] = [F] = M L T^-2. Thus, [Pd] = (M L T^-2) · L · C^-1 = M L^2 T^-2 C^-1. Step 3: Dimensions of charge [C] = [A · s] = I T. Substitute: [Pd] = M L^2 T^-2 · (I T)^-1 = M L^2 T^-2 · I^-1 T^-1 = M L^2 T^-3 I^-1. Step 4: Confirm from work per unit charge Work done = F · r, so [work] = (M L T^-2) · L = M L^2 T^-2. [Pd] = [work][q] = M L^2 T^-2I T = M L^2 T^-3 I^-1. Step 5: E.M.F. has same dimensions as electric potential difference. [Pd] = M L^2 T^-3 I^-1