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ivolga24 [154]
3 years ago
12

How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 4.57×10−21 n

ewtons?
Physics
1 answer:
hichkok12 [17]3 years ago
5 0

Answer:

891 excess electrons must be present on each sphere

Explanation:

One Charge = q1 = q

Force = F = 4.57*10^-21 N  

Other charge = q2 =q

Distance = r = 20 cm = 0.2 m  

permittivity of free space = eo =8.854×10−12 C^2/ (N.m^2)  

Using Coulomb's law,

F=[1/4pieo]q1q2/r^2

F = [1/4pieo]q^2 / r^2

q^2 =F [4pieo]r^2

q =  r*sq rt F[4pieo]

 q=0.2* sq rt[ 4.57 x 10^-21]*[4*3.1416*8.854*10^-12]

q = 1.42614*10^ -16 C

number of electrons = n = q/e=1.42614*10^ -16 /1.6*10^-19

n =891

 891 excess electrons must be present on each sphere  

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Answer:

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Explanation:

Using the equation of motion,

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make s the subject of the equation,

s = (v² - u²)/2a----------- Equation 2

Given: v = 6.4 m/s, u = 0 m/s ( from rest)  a = 0.380 m/s².

Substitute into equation 2

s = (6.4²-0²)/(2×0.380)

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<h2>Given that,</h2>

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