Asked by VJC
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?
Answers
Answered by
drwls
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.
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.
Answered by
VJC
what is H?
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