Physics, 24.07.2019 14:30 lexhorton2002

4. complete each of the following parts and show your work: a.) draw the electric field around a +2.0 × 10–5 c point source charge. draw the field lines and indicate their direction. draw two different equipotential lines. b.) calculate the force between a +5 µc test point charge and this source charge at a distance of 2.00 cm. (µc = 1.0 × 10–6 c) c.) if the test charge were moved closer to the source charge, would the change in its potential energy be positive, negative or zero? explain.

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Answer from: xojade

(a) the electric field generated by a single point charge is radial, and its direction is determined by the sign of the source charge. In particular, when the source charge is positive as in this example, the field lines are directed away from the charge (see attached figure).

(b) The electric field strength at distance r from the source charge is given by
$E=k_e \frac{Q}{r^2}$
where ke is the Coulomb's constant and Q is the source charge. In our problem, the distance is r=2.00 cm=0.02 m, so the electric field at that point is
$E=(8.99 \cdot 10^9 N m^2 C^{-2}) \frac{2.0 \cdot 10^{-5}C}{(0.02 m)^2}=4.5 \cdot 10^8 N/C$

The force experienced by the test charge in this electric field is equal to
F=qE

where $q=+5 \mu C=+5 \cdot 10^{-6}C$ is our test charge. Substituting, we find that the force is equal to
$F=qE=(5\cdot 10^{-6}C)(4.5 \cdot 10^8 N/C)=2250 N$

(c) The electric potential at a distance r from the source (positive) charge is given by
$V=k_e \frac{Q}{r}$
and the electric potential energy of a test charge q at distance r from the source charge is
$U=qV=k_e \frac{qQ}{r}$
We can see that when the test charge q is moved closer to the source Q, the distance r decreases, and so the potential energy U of the charge increases. So, the change in potential energy is positive.
4. complete each of the following parts and show your work: a.) draw the electric field around a +2

4. complete each of the following parts and show your work: a.) draw the electric field around a +2

Answer from: Quest

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

Answer from: Quest

the answer is neither ball has a charge.

because if they were both positively charged, they will repel each other.

if they were positively and negatively charged, they will seek.

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4. complete each of the following parts and show your work: a.) draw the electric field around a +2...

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