A gold nucleus (radius r) is represented by the symbol 19779Au Taking e as the elementary charge and εo as the permittivity of free space, what is the electric field strength at the surface of an isolated gold nucleus?
Correct answer: C. 79e/4 π εo r2
- A. zero
- B. 197e/(4 π εo r2 )
- C. 79e/4 π εo r2
- D. 79e2/(4 π εo r2 )
Explanation
The electric field strength E at the surface of a charged sphere can be calculated using the formula: E = kQ/r2 where:E is the electric field strength, k is Coulomb's constant ,Q is the charge on the sphere, and r is the radius of the sphere. In the case of a 19779Au ( a gold nucleus) with charge Q , the charge is equal to the number of protons n 79 multiplied by the charge on a proton e.i.e. Q=ne =79eThe formula becomes: E = k(79e)/r2 = 79e/(4 π εo r2) As, k= 1/(4 π εo )
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