Laboratory 4-Electrostatics Part 3 (1) Diameter of a pith ball d ± d (units) 10.

Question

Laboratory 4-Electrostatics Part 3 (1) Diameter of a pith ball d ± d (units) 10.5) (2) Calculation of the surface area of the pith ball (1 A-4(o.ooko) o.ss A (units) 10.25) 6 0335 m (3) Calculation of the approximate number of atoms on the surface of the pith ball (0.5) Number of atoms on the surface of the pith ball (units) 0.25) 4.8S 10 (4) Calculation of number of excess electrons if qg -2.13 C. (O.s Excess electrons (units) 0.25 Ratio atoms/excess electrons [0.25) (5) Calculation of the ratio of atoms on the surface to the number of excess electrons

Explanation / Answer

4ans)

charge of electron e=1.6x10-19 C

charge q=ne

n=q/e

given q=-2.13C

n=(-2.13)/(1.6x10-19)

n=-1.33x1019

5ans)

Already calculated in 3rd answer

Atoms on the surface =7.55x1018

Ratio of atoms on the surface to the number of excess electrons is

Number of excess electrons n=-1.33x1019

Ratio=(7.55x1018)/(-1.33x1019)

= -0.567

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