This is false. Current is the speed of the charge, 1 amp of current is 1 coulomb per second. So you can imagine the current of a circuit as the current of a river. In a parallel circuit, the river breaks into two separate streams. Some of the water goes down one river, some goes down the other. However, the total amount of water/coulombs never changes. This means that some of the total current will go down one river, and one the other. However, with less coulombs now the current will decrease.
Long story short, since there are two paths, the charge will split and depending on the resistance of each parallel stream a different amount of charge will go down each branch.
Answer:
its like to orginized your stuff and at the end have a answer
Answer:
Zinc- Zn
Mass Numbers:
Tin-118.71 u
Zinc- 65.38
Protons:
Tin-50
Atomic number zinc 45.38
Zinc has 30 electrons
Tin has 50 electrons
35 neutrons in zinc
iodine-
atomic number- 53
78 neutrons
cl-1 is chlorine
atomic number 17
18 neutrons
17 electrons
atomic number 35 is bromine
symbol Br
35 protons
45 neutrons
35 electrons
Aluminium
protons and electrons 13
neutrons 14
atomic number 14
fe+2
26 protons
24 electrons
30 neutrons
ferrous iron
Cu+1 copper
63.546 mass number
proton and electron 29
neutron 35
<h2>
Answer:</h2>
(a) 6.95 x 10⁻⁸ C
(b) 6.25N/C
<h2>
Explanation:</h2>
The electric field (E) on a point charge, Q, is given by;
E = k x Q / r² ---------------(i)
Where;
k = constant = 8.99 x 10⁹ N m²/C²
r = distance of the charge from a reference point.
Given from the question;
E = 10000N/C
r = 0.250m
Substitute these values into equation(i) as follows;
10000 = 8.99 x 10⁹ x Q / (0.25)²
10000 = 8.99 x 10⁹ x Q / (0.0625)
10000 = 143.84 x 10⁹ x Q
Solve for Q;
Q = 10000/(143.84 x 10⁹)
Q = 0.00695 x 10⁻⁵C
Q = 6.95 x 10⁻⁸ C
The magnitude of the charge is 6.95 x 10⁻⁸ C
(b) To get how large the field (E) will be at r = 10.0m, substitute these values including Q = 6.95 x 10⁻⁸ C into equation (i) as follows;
E = k x Q / r²
E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 10²
E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 100
E = 6.25N/C
Therefore, at 10.0m, the electric field will be just 6.25N/C
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