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dimension of potential difference
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(Original post by bijay44)
dimension of potential difference
Hello and welcome to TSR.

I assume you mean you want to know what the base units for potential difference are?

This is quite a nice question and not so difficult if you take things step by step:

To do this, start with the derived units of pd which is measured in Volts. and the definition of voltage is Joules/Coulomb of charge (JC-1) or V = E/Q

Now break down E and Q. Where you meet a derived unit, break it down further until you get to the base SI units and can go no further.

At that point replace the equations with their appropriate SI base units and rationalise where possible.


V = E/Q = JC-1 (both Joules and Coulombs are derived units so need to be broken down further)

Joules are the derived units of energy where:

E = F x d (distance is measured in the base unit of metres so cannot be reduced. But force is measured in Newtons which is a derived unit)

Force = Mass x Acceleration (mass is measured in the base unit of kg so cannot be reduced.)

Acceleration = rate of change of velocity and velocity is the rate of change of distance. So:

Acceleration = v/t = (distance/time)/time = m.s-2 (these are base units so cannot be reduced further)

So Energy = Force x Distance
= (Mass x Acceleration) x Distance = (kg.m.s-2)m = kg.m2s-2

Similarly Charge (Coulombs) = Current x Time = A.s (these are base units and canit be reduced)

Putting the two parts together:

Voltage = Energy / Charge

Voltage = kg.m2.s-2 / A.s

Voltage = kg.m2.A-1.s-3

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