Let the rod be on the x-axes with endpoints -L/2 and L/2 and uniform charge density lambda (2.6nC/0.4m = 7.25 nC/m). 
The point then lies on the y-axes at d = 0.03 m. 
from symmetry, the field at that point will be ascending along the y-axes. 
A charge element at position x on the rod has distance sqrt(x^2 + d^2) to the point. 
Also, from the geometry, the component in the y-direction is d/sqrt(x^2+d^2) times the field strength. 
All in all, the infinitesimal field strength from the charge between x and x+dx is: 
dE = k lambda dx * 1/(x^2+d^2) * d/sqrt(x^2+d^2) 
Therefore, upon integration, 
E = k lambda d INTEGRAL{dx / (x^2 + d^2)^(3/2) } where x goes from -L/2 to L/2. 
This gives:
E = k lambda L / (d sqrt((L/2)^2 + d^2) ) 
But lambda L = Q, the total charge on the rod, so 
E = k Q / ( d * sqrt((L/2)^2 + d^2) )
        
             
        
        
        
5 meteals that are more useful for electroplating carbon,seth,
        
             
        
        
        
Answer:
-20,000N
Explanation:
Force (N) = mass (kg) x acceleration (m/s²)
So,
Force = 2000 x -10 
 = -20,000N (Newtons)
 
        
             
        
        
        
Answer:
The center of mass of the two-ball system is 7.05 m above ground.
Explanation:
<u>Motion of 0.50 kg ball:</u>
Initial speed, u = 0 m/s
Time = 2 s
Acceleration = 9.81 m/s²
Initial height = 25 m
Substituting in equation s = ut + 0.5 at²
                  s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m
Height above ground = 25 - 19.62 = 5.38 m
<u>Motion of 0.25 kg ball:</u>
Initial speed, u = 15 m/s
Time = 2 s
Acceleration = -9.81 m/s²
Substituting in equation s = ut + 0.5 at²
                  s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m
Height above ground = 10.38 m
We have equation for center of gravity
           
           m₁ = 0.50 kg
           x₁ = 5.38 m        
           m₂ = 0.25 kg
           x₂ = 10.38 m    
Substituting
          
 The center of mass of the two-ball system is 7.05 m above ground.  
 
        
             
        
        
        
In the circuit outside of the battery the electrons have to expend all of their energy on the internal resistance of the battery which causes heating