On planet Tehar, the free-fall acceleration is the same as that on Earth, but th

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Question

On planet Tehar, the free-fall acceleration is the same as that on Earth, but there is also a strong downward electric field that is uniform close to the planet's surface. A 1.86-kg ball having a charge of 4.90 C is thrown upward at a speed of 20.1 m/s. It hits the ground after an interval of 4.10 s. What is the potential difference between the starting point and the top point of the trajectory? (Use 9.8 m/s2 for the acceleration due to gravity.) On planet Tehar, the free-fall acceleration is the same as that on Earth, but there is also a strong downward electric field that is uniform close to the planet's surface. A 1.86-kg ball having a charge of 4.90 C is thrown upward at a speed of 20.1 m/s. It hits the ground after an interval of 4.10 s. What is the potential difference between the starting point and the top point of the trajectory? (Use 9.8 m/s2 for the acceleration due to gravity.)

Explanation / Answer

mass=1.86-kg

charge = 4.90 C

speed = 20.1 m/s.

time= 4.10 s.

So the electric field is Upwards for the positive charge

20.1 = a .(4.10/2)

a = 9.80 m/s/s

We know that,

E . e = M ( g – a )=M(9.81-9.80) = 1.86*0.005 =0.0095 N

E = Force/charge=0.0095 / 4.9* 10–6 = 1944 V/m

Height reached,h = V2 / 2a = 20.12 / 2*9.8 = 20.61 m

PD = E*h=40071 Volts

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On planet Tehar, the free-fall acceleration is the same as that on Earth, but th

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