Answer:
The permittivity of rubber is 
Explanation:
From the question we are told that
The magnitude of the point charge is 
The diameter of the rubber shell is 
The Electric field inside the rubber shell is 
The radius of the rubber is mathematically evaluated as
Generally the electric field for a point is in an insulator(rubber) is mathematically represented as
Where 
is the permittivity of rubber
=> 
=> 
substituting values
Let at any instant of time the speed is vo and the angle made by the bike with the horizontal is given
now we have
component of speed in x direction given as
component of speed in y direction will be
now from above two equations we can say that here
= angle with the horizontal at any instant
and since here it is a sine curve so we know that
so we have slope of graph
First do 1.6 m (how far he jumps) 9.8 m/s (what gravity is measured at) then times 2
= 31.36
Sq root = 5.6
We have all the charges for q1, q2, and q3.
Since k = 8.988x10^2, and N=m^2/c^2
F(1) = F (2on1) + F (3on1)
F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 | / (.2m)^2
F(2on1) = 3.37 N
Since F1 is 7N,
F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)
Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N
F(3on1) = k |q1 q3| / r(the distance between the two)^2
r^2 x F(3on1) = k |q1 q3|
r = sqrt of k |q1 q3| / F(3on1)
= .144 m (distance between q1 and q3)
0 - .144m
So it's located in -.144m
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