Introduction to Electric Circuits

1.3 Dimensional analysis 5

Example 1.

The force between two charges ql and q2 separated by a distance d in a vacuum
is given by F = qlq2//4"rrEo d2, where e0 is a constant whose dimensions are
[M L -3 T 4 m2]. Check the dimensional balance of this equation.

Solution

The left-hand side of the equation is simply the force F and from Example 1.

(2) we see that its dimensions are [M L T-2].

The dimensions of the right-hand side are [q][q]/[4][Tr][E0][d2]. From Exam-

ple 1.3 (1) we see that the dimensions of electric charge [q] are [A T]. Numbers

are dimensionless so that the figure 4 and the constant 7r have no dimensions.

We are told in the question that the dimensions of E0 are [M -1 L-3T 4 m2]. The

distance between the charges, d, has the dimensions of length so that d 2 has the

dimensions [L2]. The dimensions of the right-hand side of the equation are

therefore

[A TI[A T]/[M -1 n -3 T 4 A2I[L 21 -- [A 2 TZl[M L 3 T -4 A -2 L -21 = [M L T -2]

which is the same as that obtained for the left-hand side of the equation. The

equation is therefore dimensionally balanced.

Example 1.

The energy in joules stored in a capacitor is given by the expression (CVb)//2,

where C is the capacitance of the capacitor in farads and V is the potential

difference in volts maintained across its plates. Use dimensional analysis to

determine the values of a and b.

Solution

We have that W- (cavb)/

In dimensional terms [w] = [c]a[pd] b

From Equation (1.5), [w] = [M L 2 T -2]

From Equation (1.11), [c] = [M -1 L -2 T 4 A 2]

From Equation (1.8), [pd] -- [M L 2 T -3 A -i] therefore

[M L 2 T-21 = [M -1 L -2 T 4 A2]a[M L 2 T -3 A-I] b

Equating powers of [M], 1 - -a + b ~ b - a + 1

Equating powers of [A], 0 = 2a - b ~ b = 2a

By substitution, 2a - a + 1, so

a=l and b=2a=

The values required are therefore a - 1; b - 2