Answer:
u_o = 12.91 m/s
Explanation:
Given:
- mass of the ball m = 0.15 kg
- Spacing between each equipotential line s = 1.0 m
- Charge on the ball is q = 10 mC
Find:
- What initial velocity must it have at the 200-V level for it to reach its maximum height at the 700-V level
Solution:
- Notice that the Electric field lines are directed from Higher potential to lower potential i.e V = + 700 level to V = 200 level. The gravitational acceleration also acts downwards.
- We will compute a resultant acceleration due to gravity and Electric Field as follows:
g' = g + F_e / m
- Where F_e is the Electrostatic Force given by:
F_e = E*q
- Electric field strength is given by:
E = V / d
- Hence,
g' = g + (dV*q/ d*m)
- Input values:
g' = 10 + (500*10^-2 / 5*0.15)
g' = 16.6667 m/s^2
- Now use one of the equation of motions in y-direction:
h_max = u_o ^2 / 2*g'
- Input values where h_max = 5 m
5*2*g' = u_o ^2
u_o = sqrt (166.667)
u_o = 12.91 m/s
