Answer:
The potential difference between the starting point and the top of the trajectory is 82.51V
Explanation:
Finding acceleration of the ball
Using kinematics we can find the acceleration of the ball.
[tex] v = v_{0} + at [/tex]
At the top of the trajectory we have vertical velocity v = 0 and the time is half of the total time
[tex] 0 = v_{0} + a \frac{t}{2} [/tex]
[tex] a = \frac{-2v_{0}}{t} [/tex]
[tex] a = \frac{-2 (20.1m/s)}{4.10s} [/tex]
[tex] a = 9.8049m/s^{2}[/tex]
Electric field potential difference
We can obtain the potential difference from the electric field and the distance between the starting point and the top of the trajectory by using the following equation:
[tex] V= \frac{E}{d} [/tex]
d being the distance from the surface to the top of the trajectory of the ball.
With the given values, we can find the sum of the gravitational and electric forces on the object in order to obtain the electric field on the surface of the planet.
We start with:
[tex] F_{net}=ma [/tex]
[tex] mg + qE = ma [/tex]
[tex] qE = ma - mg [/tex]
[tex] E = \frac{m}{q}(a-g) [/tex]
[tex] E = \frac{(1.84 kg)}{5.28 \mu C}(9.8049m/s^{2}- 9.8m/s^{2})[/tex]
[tex] E= 1.70 x10^{3} Vm [/tex]
Now we can calculate d using kinematics and the fact that [tex]v = 0[/tex] and [tex]a= -2v_o/t[/tex] at the top of the trajectory.
[tex] v^{2} = v_{0}^{2} + 2ad [/tex]
[tex] 0 = v_{0}^{2} + 2 (\frac{-2vo}{t}) d [/tex]
[tex] \frac{4v_{0}}{t} d = v_{0}^{2} [/tex]
[tex] d = \frac{1}{4}v_{0} t [/tex]
[tex] d= \frac{1}{4}(20.1m/s) (4.10s)[/tex]
[tex] d= 20.60m[/tex]
The last step is to use V = E/d to find the final answer:
[tex] V= \frac{E}{d} [/tex]
[tex] V= \frac{1.70 x10^{3} Vm}{20.60m}[/tex]
[tex] \boxed{V = 82.51 V} [/tex]
The potential difference between the starting point and the top of the trajectory is 82.51V
