Answer: So if you had 570 cm of ribbon, then 570%2F8.5=67.05 which means that about 67 students can do the experiment (round down to the nearest whole number).
Explanation: If you had 8.5 cm of ribbon, then only 8.5%2F8.5=1 student can do the experiment. If you had 17 cm of ribbon, then 17%2F8.5=2 students can do the experiment.
Answer:
the complete question is found in the attachment
Explanation:
the complete explanation is found in the attachment
Answer:
3.65 x 10¹⁰ electrons
Explanation:
we'll apply the following equation for electric field of a point charge on a spherical conductor
where E is the electric field
k is a constant of the value 8.99 x 10⁹ Nm²/C²
r is the radius of the spherical conductor
q is the total charge in the sphere
Given diameter d =41.0cm, radius r = 20.5cm = 0.205m (convert cm to m)
Electrical field E = 1250 N/C
we are asked to determine how many excess electrons must be added to the surface of the sphere to produce this electric field
q = <u>E x r²</u>
k
q = <u>1250 N/C x 0.205m</u>²
8.99 x 10⁹ Nm²/C²
q = 5.84 x 10⁻⁹ C
this is the total charge in the sphere
To determine the number of electrons, we can divide the charge q by the charge on an electron e (1.6 x 10⁻¹⁹C)
n = <u>5.84 x 10⁻⁹ C </u>
1.6 x 10⁻¹⁹C
n = 3.65 x 10¹⁰ electrons
Therefore, to apply an electric field of magnitude 1250 N/C, the isolated spherical conductor must contain 3.65 x 10¹⁰ electrons