Answer:
(a) 135 kV
(b) The charge chould be moved to infinity
Explanation:
(a)
The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

where 
Difference in potential between the points is
![kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}](https://tex.z-dn.net/?f=kQ%5Cleft%5B-%5Cdfrac%7B1%7D%7B0.2%5Ctext%7B%20m%7D%7D%20-%5Cleft%28%20-%5Cdfrac%7B1%7D%7B0.1%5Ctext%7B%20m%7D%7D%5Cright%29%5Cright%5D%20%3D%20%5Cdfrac%7BkQ%7D%7B0.2%5Ctext%7B%20m%7D%7D%20%3D%20%5Cdfrac%7B9%5Ctimes10%5E9%5Ctext%7B%20F%2Fm%7D%5Ctimes3%5Ctimes10%5E%7B-6%7D%5Ctext%7B%20C%7D%7D%7B0.2%5Ctext%7B%20m%7D%7D)

(b)
If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.
![270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]](https://tex.z-dn.net/?f=270%5Ctimes10%5E3%20%3D%20kQ%5Cleft%5B-%5Cdfrac%7B1%7D%7Bx%7D-%5Cleft%28-%5Cdfrac%7B1%7D%7B0.1%5Ctext%7B%20m%7D%7D%5Cright%29%5Cright%5D)



The charge chould be moved to infinity
Answer:
F = 2,894 N
Explanation:
For this exercise let's use Newton's second law
F = m a
The acceleration is centripetal
a = v² / r
Angular and linear variables are related.
v = w r
Let's replace
F = m w² r
The radius r and the length of the rope is related
cos is = r / L
r = L cos tea
Let's replace
F = m w² L cos θ
Let's reduce the magnitudes to the SI system
m = 101.7 g (1 kg / 1000g) = 0.1017 kg
θ = 5 rev (2π rad / rev) = 31,416 rad
w = θ / t
w = 31.416 / 5.1
w = 6.16 rad / s
F = 0.1017 6.16² 0.75 cos θ
F = 2,894 cos θ
The maximum value of F is for θ equal to zero
F = 2,894 N
Answer:
at point A the ball possess pontetial energy , point B kinetic energy then point C pontetial energy