The ball's acceleration downward is given by
a * (3.63/2 s) = 21.0 m/s
a = 11.57 m/s^2
Since g = 9.81 m/s^2, the downward acceleration due to electrostatic force is 1.76 m/s^2
The potential difference V due to electrostatic force at the top of the trajectory is given by
M*1.76* H = Q*V
Solve for V.
On planet Tehar, the acceleration of gravity is the same as that on Earth but there is also a strong downward electric field with the field being uniform close to the planet's surface. A 2.16 kg ball having a charge of 4.80 uC is thrown upward at a speed of 21.0 m/s and it hits the ground after an interval of 3.63 s. What is the potential difference between the starting point and the top point of the trajectory?
2 answers
what is H?