At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 V and respectively. (Take the potential to be zero at infinity.) (a) What is the distance to the point charge? (b) What is the magnitude of the charge? (c) Is the electric field directed toward or away from the point charge

(a) 0.3074 m

(b) 0.17 nC

(c) The electric field is directed away from the charge.

Explanation:

Parameters given:

Electric Potential, V = 4.98V

Electric field magnitude, E = 16.2V/m

(a) The electric potential is given as:

V = kq/r

Where

k = Coulombs constant

q = charge

r = distance

Making q the subject of formula:

q = (V*r)/k 1

Electric field magnitude, E is given as:

E = kq/r²

Making q subject of the formula:

q = (E*r²)/k 2

Equating 1 and 2,

(V*r)/k = (E*r²)/k

=> r = V/E

r = 4.98/16.2

r = 0.3074 m

(b) Charge, q, from 1:

q = (V*r)/k

q = (4.98 * 0.3074)/(9 * 10^9)

q = 0.17 nC

(c) The electric field and charge have positive values, hence the electric field is directed away from the charge.

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂