a charged van de graaf generator of radius 0.15m has a charge of about -23*10^-6C. it is then turned off and grounded. how many surplus electrons if any remain on its dome?
To determine the number of surplus electrons on the dome of the Van de Graaff generator, we can use the formula:
Q = Ne
where:
Q is the charge on the dome,
N is the number of surplus electrons, and
e is the elementary charge which is approximately equal to 1.6 x 10^-19 Coulombs.
Given:
Q = -23 x 10^-6 C (charge on the dome)
Substituting the values into the formula:
-23 x 10^-6 C = N x (1.6 x 10^-19 C)
Now, solve for N:
N = (-23 x 10^-6 C) / (1.6 x 10^-19 C)
N = -23 x 10^-6 C / 1.6 x 10^-19 C
N = -14.375 x 10^13
Since the number of electrons cannot be negative, we take the absolute value:
N ≈ 14.375 x 10^13
So, there are approximately 14.375 x 10^13 surplus electrons on the dome of the Van de Graaff generator.