(b) 
We can actually solve first part (b) the problem. In fact, we know that the electric field strength at the surface of the a sphere is given by
where
k is the Coulomb's constant
Q is the charge on the surface of the sphere
R is the radius
For this sphere, the radius is half the diameter, so
We also know that the maximum charge is the value of charge deposited at which the electric field of the sphere becomes equal to the breakdown electric field, so
Solving the formula for Q, we find the maximum charge:
(a) 
The maximum potential of a charged sphere occurs at the surface of the sphere, and it is given by
where we already found at point b)
and we know that
R = 0.215 m
Solving for V, we find:
Answer:
(a) 27.1 m/s
(b) 3.9 second
Explanation:
Let the speed is u.
Maximum horizontal range, R = 75 m
The range is maximum when the angle of projection is 45°.
(a) Use the formula for the maximum horizontal range
u = 27.1 m/s
(b) Let the time of flight is T.
Use the formula for the time of flight
T = 3.9 second
<span>If an electrostatically charged object is placed near other objects, </span>it would be attracted to objects with an opposite charge.
Answer:
The distance of separation is decreased
Explanation:
From Cuolomb's law, we know that the strength of charge is inversely proportional to the distance of separation between the charges. To mean that increasing the distance let's say from 2m to 3 m would mean initial strength getting form 1/4 to 1/9 which is a decrease. The vice versa is true hence the force of repulsion can increase only when we decrease the distance of separation.
