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Question: The SI unit of electric potential is A.\(Kg{m^2}{s^2}C\) B.\(Kg{m^2}C{s^{ - 2}}\) C.\(Kg{m^2}...

The SI unit of electric potential is
A.Kgm2s2CKg{m^2}{s^2}C
B.Kgm2Cs2Kg{m^2}C{s^{ - 2}}
C.Kgm2A1s3Kg{m^2}{A^{ - 1}}{s^{ - 3}}
D.Kgms3C1Kgm{s^{ - 3}}{C^{ - 1}}

Explanation

Solution

Substitute the base units of volt by simplifying the SI units of other quantities in it by using the definition of potential, into their base units. For example converting newton into base units of force.

Complete answer:
The electric potential is defined as the amount of work needed to shift a unit charge from a point A to a point B. It is denoted by VV, and its SI unit is volt.

Shifting the charge from point A to B.

Formula of electric potential VV is,
V=kq/rV = kq/r; …………………..(equation 1)
Where,
k=k = coulomb constant; and its value is 9×109Nm2C29 \times {10^9}N{m^2}{C^{ - 2}} .
q=q = charge; its SI unit is coulomb and it is denoted by CC .
r=r = distance; its SI unit is meter and it is denoted by mm .

Now, by applying all the above SI units to equation 1 we get,
V=Nm2C2×Cm1V = N{m^2}{C^{ - 2}} \times C{m^{ - 1}}
By adding powers of mm and CC in above equation we get,
V=NmC1V = Nm{C^{ - 1}}; ……………………...(equation 2)
Now, we know that newton denoted as NN is the SI unit of force. And formula of force is;
F=maF = ma; …………………...(equation 3)
Where, F=F = force; its SI unit is newton and it is denoted by NN,
m=m = mass; its SI unit is kilogram and it is denote by KgKg,
a=a = acceleration; its SI unit is meter per second square and it is denoted as ms2m{s^{ - 2}}.
Now, by applying all the above SI units to equation 3 we get,
N=Kgms2N = Kgm{s^{ - 2}}; ………………….. (equation 4)
Now, by substituting equation 4 in equation 2 we get,
V=Kgms2×mC1V = Kgm{s^{ - 2}} \times m{C^{ - 1}}
By adding powers of mm we get,
V=Kgm2s2C1V = Kg{m^2}{s^{ - 2}}{C^{ - 1}}; ………………..(equation 5)
Since we know that formula of electric current is
I=q/tI = q/t ; ………………….(equation 6)
Where,
II is electric current and denoted by AA , qq is charge denoted by CC, and tt is time denoted by ss.
By applying all the above SI units to equation 6 we get,
A=Cs1A = C{s^{ - 1}}
By converting the equation with respect to CC we get,
C=AsC = As …………………..(equation7)
Now, by substituting equation 7 in equation 5 we get,
V=Kgm2s2(As)1V = Kg{m^2}{s^{ - 2}}{(As)^{ - 1}} ;
By solving powers of ss we get,
V=Kgm2A1s3V = Kg{m^2}{A^{ - 1}}{s^{ - 3}}; ……………...(equation 8)
Since we have simplify all the units to simpler units, we get our final answer as
c. V=Kgm2A1s3V = Kg{m^2}{A^{ - 1}}{s^{ - 3}};

Hence, the correct option is (C).

Note: In such questions try simplifying the SI units into simpler base units, until you get one of the given options. To avoid confusion, always try to solve powers before going to the next step.