A metal sphere of radius 19 cm has a net charge of 2.4 × 10–8 C. (a) What is the electric field at the sphere's surface? (b) If V = 0 at infinity, what is the electric potential at the sphere's surface? (c) At what distance from the sphere's surface has the electric potential decreased by 370 V?

1.29*10^6 N/C

1135.6 V

9.18 cm

Explanation:

Given that

radius of the metal, r = 19 cm

charge of the metal, q = 2.4*10^-8 C

coulomb's constant, k = 8.99*10^9

to find the electric field, we use the formula E = kq/r², where

E = electric field

k = coulomb constant

q = charge on the metal and

r = radius of the metal

E = (8.99*10^9 * 2.4*10^-8) / 0.19²

E = 215.76 / 0.0361

E = 1.29*10^6 N/C

to find the electric potential, we use this relation

V = kq/r

V = (8.99*10^9 * 2.4*10^-8) / 0.19

V = 215.76 / 0.19

V = 1135.6 V

V = kq/r,

r = kq/V

r = (8.99*10^9 * 2.4*10^-8)/ (1135.6 - 370

r = 215.76 / 765.6

r = 0.281 = 28.1 cm

distance from the sphere

28.18 - 19 = 9.18 cm

depends on what you mean by "group." for order, it would be coleoptera (beetles). for class, it would be insecta (insects). for phylum, it would be the arthropoda (animals with jointed exoskeletons).

explanation:

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form