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sergeinik [125]
3 years ago
14

How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 30.0 cmcm in diameter to

produce an electric field of 1440 N/CN/C just outside the surface of the sphere
Physics
1 answer:
svp [43]3 years ago
3 0

Answer:

1.78×10×10^10 electron

Explanation:

Electric field outside the sphere can be calculated using below expression

E= kq/ r^2..........eqn(1)

Where k= 9 × 10^9 NM^2/C^2

q= charge

E= 1440 N/C

Diameter= 30.0 cm= 0.3 m

r= radius= 0.3/2= 0.15m

If we make q subject of formula from eqn(1) we have

q= Er^2/k............eqn(2)

q= 1440 × (0.15)^2 /(9 × 10^9 )

= 2.85×10^-9C

Total charge is an integer of electron charge , then we can calculate the number of the electron using the expression below

q= Ne

Where

N = number of electron

Making N subject of the formula we have

N= q/e

Where e= electron value= 1.6× 10^-19

N=2.85×10^-9 /1.6× 10^-19

= 1.78×10×10^10 electron

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Alan starts from his home and walks 1.3 km east to the library. He walks an additional 0.68 km east to a music store. From there
zepelin [54]

Answer: final Displacement = 0 km, total distance covered =7 km

Explanation:

Given the following :

From home to library = 1.3 km East

Library to music store = 0.68km East

Music store to friend's house = 1.1km North

Friend's house to grocery store = 0.42 km North

Displacement is the net change in distance traveled.

Eastward distance :

To = (1.3 + 0.68)km = 1.98km East

Fro = (0.68 + 1.3)km = 1.98 km East

Δ distance = (1.98 - 1.98) = 0

Northward direction:

To = (1.1 + 0.42)km = 1.52km north

Fro = (0.42 + 1.1)km = 1.52km North

Δ distance = (1.98 - 1.98) = 0

Hence final Displacement = 0

Total distance covered = 2 × (1.3 + 0.68 + 1.1 + 0.42) = 2 × 3.5

= 7km

3 0
4 years ago
What are basic physical quantities and it's measurement units? Make a chart.
jolli1 [7]

Displacement/distance metres
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8 0
3 years ago
A magnetic field of 37.2 t has been achieved at the mit francis bitter national magnetic laboratory. Find the current needed to
jeka57 [31]

Answer:

Here is the complete question:

https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632

a) Current for long straight wire  =3.7\ MA

b) Current at the center of the circular coil =2.48\times 10^{5}\ A

c) Current near the center of a solenoid 236.8\ A

Explanation:

⇒ Magnetic Field due to long straight wire is given by (B),where B=\frac{\mu\times I}{2\pi(r) },so\ I=\frac{B\ 2\pi(r)}{\mu}

\mu=4\pi \times 10^{-7}\ \frac{henry}{m}

Plugging the values,

Conversion 1\times 10^6 A = 1\ MA,and 2cm=\frac{2}{100}=0.02\ m

I=\frac{37.2\times \ 2\pi(0.02)}{4\ \pi \times (10^{-7})}=3.7\ MA

⇒Magnetic Field at the center due to circular coil (at center) is given by,B=\frac{\mu\times I (N)}{2(a)}

So I= \frac{2B(a)}{\mu\ N} = \frac{2\times 37.2\times 0.42}{4\pi\times 10^{-7}\times 100}=2.48\time 10{^5}\ A

⇒Magnetic field due to the long solenoid,B=\mu\ nI=\mu (\frac{N}{l})I

Then I=\frac{B}{\mu(\frac{N}{L})} \approx 236.8\A  

So the value of current are  3.7 MA,2.48\times 10^{5} A and 236.8\ A respectively.

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