Answer:
n = 1/5 and m = 3/5
Explanation:
The given quantity is :
Where
The dimension of [A] = [LT]
The dimension of [B] = [L²T⁻¹]
The dimension of [C] = [LT²]
We need to find the dimensions of n and m values.
Using dimensional analysis,
Comparing both sides,
2n+m=1 ....(1)
-n+2m=1 ,.....(2)
Solving (1) and (2), we get :
n = 1/5 and m = 3/5
Hence, this is the required solution.
Answer:
1.55 m
Explanation:
The potential produced by a point charge, is inversely proportional to the distance from the charge to the point where the potential is being calculated, as follows:
As it only depends from the distance r, we can conclude that if the potential is the same for any point to a distance r from the point charge, the equipotencial surface must be a sphere of radius r.
Replacing q = +1.7*10⁻⁸ C, and k = 9*10⁹ N*m²/C², and V, by 120 V and 54 V, we can find the distance from the charge, to the points where we are calculating the potential V, as follows:
The distance between both points, is just the difference between the radius of both spheres, as follows:
r₂ - r₁ = 1.55 m