Answer:
Part A : E = 
ε₀ Q₁/R₁² Volt/meter
Part B : V = ε₀ Q₁/R₁ Volt
Explanation:
Given that,
Charge distributed on the sphere is Q₁
The radius of sphere is R
₁
The electric potential at infinity is 0
<em>Part A</em>
The space around a charge in which its influence is felt is known in the electric field. The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.
If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by
F = ε₀ Q₁/R₁²
Then the electric field at that point is
E = F/1
E = ε₀ Q₁/R₁² Volt/meter
Part B
The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.
Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation
V = ε₀ Q₁/R₁ Volt
Answer:
(a) Height is 4.47 m
(b) Height is 4.37 m
Solution:
As per the question:
Initial velocity of teh ball, 
Angle made by the ramp, 
Distance traveled by the ball on the ramp, d = 5.00 m
Now,
(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:
where
H = 
g = 
= 19.06 m/s
Now, maximum height attained is given by:
Height from the ground = 
(b) now, considering the coefficient of friction bhetween ramp and the ball, 
:
velocity can be given by the eqn-3 of motion:
= 18.7 m/s
Now, maximum height attained is given by:
Height from the ground = 
The potential energy is most often referred to as the "energy at rest" and is dependent on the elevation of an object. This can be calculated through the equation,
E = mgh
where E is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height. In this item, we are not given with the mass of the cart so we assume it to be m. The force is therefore,
E = m(9.8 m/s²)(0.5 m) = 4.9m
Hence, the potential energy is equal to 4.9m.
