The cart comes to rest from 1.3 m/s in a matter of 0.30 s, so it undergoes an acceleration <em>a</em> of
<em>a</em> = (0 - 1.3 m/s) / (0.30 s)
<em>a</em> ≈ -4.33 m/s²
This acceleration is applied by a force of -65 N, i.e. a force of 65 N that opposes the cart's motion downhill. So the cart has a mass <em>m</em> such that
-65 N = <em>m</em> (-4.33 m/s²)
<em>m</em> = 15 kg
Electric field strength near the surface of sphere is given by
Now we have
Q = 1 C
R = 5 cm
now electric field is given by
Part B)
Since charge on the sphere is very large to electric field near its surface is too high. So this is reasonable result near it.
Part c)
As per our assumption the charge is uniformly distributed on the surface and there is no effect of surrounding charge or surrounding electric field.
B. picking up a box off the floor
Answer: Magnitude of electric field =p × Ke × Q/d^3
Explanation: Using Coulombs law of point charge,each charge on the circle would exert a fieldEc at point given by:
Ec= Ke × (Q/n)/d^2
Where Ke= Coulomb's constant
d= distance between the charges and the point of measurement, P with d^2=a^2+p^2
(Q/n)= Magnitude of the charge.
For charges in a circle,all the force components and direction(x,z) are cancelled by the symmetry,leaving only the vertical force(y-direction)
The Resultant vector will be
Ecy=Eq × sin(theta)
Ecy=Ke × (Q/n)/d^2 ×(p/d)
Adding the forces from all the charges,the magnitude of electric field Ey=n×Ecy
/Ey/= n ×[Ke ×(Q/n)/d^2 ×(p/d)]
/Ey/= p × Ke × (Q/d^3)
