An equipotential surface that surrounds a point charge q has a potential of 536 V and an area of 1.20 m2. Determine q.

q = 1.84×10^-8coulombs

Explanation:

Surface area = 4πr²

r is the distance

1.2 = 4(3.14)r²

1.2 = 12.56r²

r² = 1.2/12.56

r² = 0.0956

r = √0.0956

r = 0.309m

Get the charge C

V = kq/r

536 = 9.0×10^9q/0.309

536×0.309 = 9×10^9q

165.73 = 9×10^9q.

q = 165.73/9×10^9

q = 1.84×10^-8coulombs

part 1: which substance has a higher solubility?

a

smallest amount left undissolved atall temps in table shown most has dissolved high sol'by