A metal sphere with radius 1.00 cm is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius rb = 9.40 cm. Charge +q is put on the inner sphere and charge −q on the outer spherical shell. The magnitude of q is chosen to make the potential difference between the spheres 470 V, with the inner sphere at higher potential.

Calculate q.

q = 5.84*10^-10 C

Explanation:

Given

Radius of the metal sphere, r(a) = 1 cm = 1*10^-2 m

Radius of the spherical shell, r(b) = 9.4 cm = 9.4*10^-2 m

Potential difference between the spheres, ΔV = 470 V

The potential difference of an inner sphere is given by

V(in) = kq/r(a) - kq/r(b)

while the potential difference of the outer sphere is given as

V(out) = 0

ΔV = V(in) - V(out)

ΔV = kq/r(a) - kq(b)

ΔV = kq * [1/r(a) - 1/r(b)]

where k = 9*10^9

470 = 9*10^9 * q (1/0.01 - 1/0.094)

470 = 9*10^9 * q *(100 - 10.638)

470 = 9*10^9 * q * 89.362

q = 470 / (9*10^9 * 89.362)

q = 470 / 8.04*10^11

q = 5.84*10^-10 C

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

see earlier answer