Question

If C is capacitance and V is Potential difference then unit of CV is: plz fast​

Answer 1

Answer:

C=qV=qW/q=q2W=(AT)2ML2T−2C=qV=qW/q=q2W=(AT)2ML2T-2

[M−1L−2T4A2][M-1L-2T4A2]

V=[ML2T−3A−1]V=[ML2T-3A-1]

ρ=RaI=VIaI=(ML2T−3A−1)(L2)ALρ=RaI=VIaI=(ML2T-3A-1)(L2)AL

=[ML3T−3A−2]=[ML3T-3A-2]

From F=q1q24π∈0r2F=q1q24π∈0r2

in0=q1q24πFr2=(AT)2(MLT)−2(L2)=[M−1L−3T4A2]in0=q1q24πFr2=(AT)2(MLT)-2(L2)=[M-1L-3T4A2]

Now, CVρin0=qρin0CVρin0=qρin0

=AT(ML3T−3)(M−1L−3T4A2)=AT(ML3T-3)(M-1L-3T4A2)

= [A]

:. dimension of A = 1

Answer 2

Answer:

it is coulomb

Explanation:

according to the formula

Q = CV

here Q represents charge stored in capacitor and hence unit of charge is coulomb

dimensionally,

C = Aε₀ / d

where a is area

d is distance between the plates

and ε₀ is permitivity of free space

C =[ L² ] [M⁻¹ L⁻³ T⁴ A² ]/  [L]

  = [ M⁻¹ L⁻² T⁴ A²]

V= [ M L² T⁻³ A⁻¹]

WHEN MULTIPLIED

CV = [ AT]

which is dimensionally charge and hence unit is coulomb

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