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Tom [10]
3 years ago
8

Identify the areas on the image where the force of repulsion is the least.

Physics
2 answers:
timofeeve [1]3 years ago
7 0

Answer:

here in the first figure the force of repulsion will be least between two SOUTH poles at the ends of the magnet

while in the other diagram the repulsion force of left end or repulsion force of each right end will be least on each other

Explanation:

the force of repulsion is always between two similar type of poles between two magnets

So when similar poles are close to each other then it will have more repulsion force while if the similar poles are far apart then the repulsion force will be smaller

So here in the first figure the force of repulsion will be least between two SOUTH poles at the ends of the magnet

while in the other diagram the repulsion force of left end or repulsion force of each right end will be least on each other

ArbitrLikvidat [17]3 years ago
3 0

Each magnet has a north pole and a south pole. We know that, from having played with bar magnets in our childhood, that a magnet's north pole will repel another magnet's north pole and attract its south pole.

From this diagram it is easy to see that the two lower bar magnets not only repel each other, but they are quite attracted to each other since their north and south poles are close together.

Therefore the region between the lower two magnets has the least force of repulsion.

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A solid sphere of radius 40.0 cm has a total positive charge of 16.2 μC uniformly distributed throughout its volume. Calculate t
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Answer:

(a) E=0  :   0 cm from the center of the sphere

(b) E= 227.8*10³ N/C   :    10.0 cm from the center of the sphere

(c)E= 911.25*10³ N/C    :    40.0 cm from the center of the sphere

(d)E= 411.84 * 10³ N/C  :    59.5 cm from the center of the sphere

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

E=\frac{K*Q}{r^{2} } : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

E=k*\frac{Q}{a^{3} } *r :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q:  Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at  0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1

E= 227.8*10³ N/C

(c) Magnitude of the electric field at 40.0 cm from the center of the sphere

r=a, We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }

E= 911.25*10³ N/C

(d) Magnitude of the electric field at 59.5 cm from the center of the sphere  

r>a , We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }

E= 411.84 * 10³ N/C

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