Question

Use Coulomb’s law to derive the dimension for the permittivity of free space. ​

Answer 1

Answer: M^-1 L^-3T^4A^2

Explanation:

From coloumb's law

K = q1q2 / (F × r^2)

Where;

q1, q2 = charges

k = constant (permittivity of free space)

r = distance

Charge (q) = current(A) × time(T) = TA

THEREFORE,

q1q2 = (TA) × (TA) = (TA)^2

Velocity = Distance(L) / time(T) = L/T

Acceleration = change in Velocity(L/T) / time (T)

Therefore, acceleration = LT^-2

Force(F) = Mass(M) × acceleration (LT^-2)

Force(F) = MLT^-2

Distance(r^2) = L^2

From ; K = q1q2 / (F × r^2)

K = (TA)^2 / (MLT^-2) (L^2)

K = T^2A^2M^-1L^-1T^2 L^-2

COLLEXTING LIKE TERMS

T^2+2 A^2 M^-1 L^-1-2

M^-1 L^-3T^4A^2