Answer:
6.78 X 10³ N/C
Explanation:
Electric field near a charged infinite plate
= surface charge density / 2ε₀
Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.
Field due to charge density of +95.0 nC/m2
E₁ = 95 x 10⁻⁹ / 2 ε₀
Field due to charge density of -25.0 nC/m2
E₂ = 25 x 10⁻⁹ / 2ε₀
Total field
E = E₁ + E₂
= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ / 2ε₀
= 6.78 X 10³ N/C
Answer:
The question has some details missing, here is the complete question ; A -3.0 nC point charge is at the origin, and a second -5.0nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = 0.200 m.
Explanation:
The application of coulonb's law is used to approach the question as shown in the attached file.
Answer:
Explanation:
Given that,
Spring constant = 16N/m
Extension of spring
x = 8cm = 0.08m
Mass
m = 5g =5/1000 = 0.005 kg
The ball will leave with a speed that makes its kinetic energy equal to the potential energy of the compressed spring.
So, Using conservation of energy
Energy in spring is converted to kinectic energy
So, Ux = K.E
Ux = ½ kx²
Then,
Ux = ½ × 16 × 0.08m²
Ux = 0.64 J
Since, K.E = Ux
K.E = 0.64 J
If i remeber correctly when dealing with real world cordinate systems as you rotate around clockwise you move in a positive direction. but all the examples i have done said north was 0 degrees, so i may be wrong
Answer:
a) 
b) 
c) Compressing is easier
Explanation:
Given:
Expression of force:

where:



when the spring is stretched
when the spring is compressed
hence,

a)
From the work energy equivalence the work done is equal to the spring potential energy:
here the spring is stretched so, 
Now,
The spring constant at this instant:



Now work done:



b)
When compressing the spring by 0.05 m
we have, 
<u>The spring constant at this instant:</u>



Now work done:



c)
Since the work done in case of stretching the spring is greater in magnitude than the work done in compressing the spring through the same deflection. So, the compression of the spring is easier than its stretching.