Answer:
F = 0.1575 N
Explanation:
When the third sphere touches the first sphere, the charge is distributed between both spheres, then now the first sphere has only half of his original charge.
In this moment then
Sphere one has a charge = Q/2
Sphere three has a charge = Q/2
Now when the third sphere touches the second sphere again the charge is distributed in a manner that both sphere has the same charge.
How the total charge is Q = Q/2 + Q = 3/2Q, when the spheres are separated each one has 3/4Q
Sphere two has a charge = 3/4Q
Sphere three has a charge = 3/4Q
The electrostatic force that acts on sphere 2 due to sphere 1 is:
F = 
F= 
how
= 0.42
Then
F = 
F = 0.1575 N
Answer: Not 100% sure but I believe the answer is B.
Hope this helps! ^^
When is at the end of the runway the velocity of the plane is given by the equation

where s=1800 m is the runway length. Thus
At half runway the velocity of the plane is

Therefore at midpoint of runway the percentage of takeoff velocity is
‰
Answer:
The electric field at origin is 3600 N/C
Solution:
As per the question:
Charge density of rod 1, 
Charge density of rod 2, 
Now,
To calculate the electric field at origin:
We know that the electric field due to a long rod is given by:

Also,
(1)
where
K = electrostatic constant = 
R = Distance
= linear charge density
Now,
In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.
At x = - 1 cm = - 0.01 m:
Using eqn (1):

(towards)
Now, at x = 1 cm = 0.01 m :
Using eqn (1):

(towards)
Now, the total field at the origin is the sum of both the fields:
