QUESTION

# The Dimensional formula for the electric field is………………………………….. $(a){\text{ }}\left[ {M{L^2}{T^{ - 3}}{A^{ - 1}}} \right] \\ (b){\text{ }}\left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right] \\ (c){\text{ }}\left[ {ML{T^{ - 3}}{A^{ - 1}}} \right] \\ (d){\text{ }}\left[ {{M^0}{L^0}{T^0}{A^0}} \right] \\$

Hint – In this question use the dimensional formula of force and charge, as the electric field can be taken as the ratio of force to charge hence the ratio of the dimensions of these two will give the required dimension.

As we know that the electric field (E) is the ratio of force (F) to charge (q)
Therefore, E = (F/q)
Now as we know force is the product of mass (M) to square of acceleration (a)
Therefore, F = (M. a2)
Now as we know that charge (q) is the product of current (I) and time (t).
Therefore, q = (I. t)
Now as we know that the dimension of current (I) is (A1) and the dimension of time (t) is (T1).
So the dimension of charge (q) is (A1. T1).
Now as we know that the dimension of mass (M) is (M1).
And we know the S.I unit of acceleration (a) is m/s2.
The dimension of meter is (L1) and the dimension of second (s) is (T1).
So the dimension of acceleration is (L1T-2).
Therefore, the dimension of force (F) is [M1 L1T-2].
Therefore, the dimension of the electric field is [M1 L1T-2]/[A1T1].
So on simplifying we get,
Dimension of electric field (E) = [M1 L1 A-1 T-3].
So this is the required dimension.
Hence option (C) is correct.

Note – Dimension formula is the expression for the unit of a physical quantity in terms of the fundamental quantities. The fundamental quantities are mass (M), Length (L) and time (T). A dimensional formula is expressed in terms of power of M,L and T.