Consider a regular pyramid charged with constant volume charge density. The potential at point P, the top of the pyramid is 1V. What would be the potential at point P if we removed the top part, so the pyramid is instead truncated at a height h=0.6H?
Hint: You don't need to do any integration, but instead can deduce the result logically.
Removing the top part is equivalent to adding to that top part a negative charge density that cancels out the charge there. The potential is then the sum of the potential from the original charge density and that of the negative charge density.
The negative charge density fills a rescaled version of the original pyramid, so to solve the problem you need to figure out how the potential behaves under a rescaling of the charge distribution.
The potential V(r) is, in general, given by:
V(r) = Integral rho(r')/|r-r'| d^3r'
Suppose we replace rho(r') by
rho(lambda r'). In this problem lambda would be equal to 5/2; t = lambda r' will range over the original pyramid The potential V' of the rescaled charge distribution is then:
V'(r) = 1/lambda^3 Integral rho(t)/|r-t/lambda| d^3t
If we introduce the variable
s = r lambda, this becomes:
1/lambda^2 Integral rho(t)/|s-t| d^3t
Now Integral rho(t)/|s-t| d^3t is the original potential at the point s = r lambda, so the potential due to the rescaled charge distribution is thus:
V'(r) = 1/lambda^2 V(lambda r)
So, the problem can be be solved as follows. We take the top to be at the origin r = 0. The potential due to the charge distribution is 1 V. Then the top part filled with the negative charge distribution is obtained by rescaling w.r.t. the origin by factor of lambda = 5/2.
Had we filled the whole pyramid with the negative charge distribution, that would have led to a potential of -1 V, therefore filling the top part with the negative charge distribution will lead to a contribution to the potential of
-(2/5)^2 V = -4/25 V
The total potential is thus 21/25 V
.84
isnt it 17/25 ?
To find the potential at point P after truncating the pyramid, we can deduce the result using logical reasoning.
First, let's understand the concept of potential in a regular pyramid charged with constant volume charge density. The potential at any point inside a symmetrical object depends on the distance of that point from the center of symmetry. In a regular pyramid, the center of symmetry is located at the apex.
In the given scenario, we have a regular pyramid with a constant volume charge density, and the potential at point P on the apex is given as 1V.
Now, when we truncate the pyramid at a height h = 0.6H, we remove a portion of the pyramid above this height. As a result, the truncated pyramid will have a smaller height but the same base area.
Since the charge density is constant throughout the pyramid, the charge removed from the top portion will be proportionate to the volume removed. Since the volume of a pyramid is directly proportional to the height, removing 60% (0.6) of the height will also remove 60% of the charge.
Since potential depends on the amount of charge, removing 60% of the charge means the potential at point P will also reduce by 60%.
Therefore, the potential at point P after truncating the pyramid at a height h = 0.6H would be 0.4V (0.6 * 1V).