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Electric energy - potential difference problem!

  • Thread starter physics213
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  • #1
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Homework Statement



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 2.00 kg ball having a charge of 5.00 µC 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?


Homework Equations



Work = (q)(E)(delta x)

q=charge, E= electric field

delta V = (delta PE)/q

V=electric potential difference

The Attempt at a Solution



I pretty sure we'd have to use the conservation of energy to determine the solution and the a physics 1 kinematic equation. I know the background of the question but I just can't figure out how the electric field comes into play for the question. Can someone please help me understand this question? Thanks a lot in advance for your help.
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
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Remember at all times on Tehar F = MA.

Determine through regular kinematics what the acceleration is on the ball.

F = m*g + q*E = m*a

So ...

q*E = m*a - m*g = m*(a - g)

This suggests that the E field (which is uniform) is

E = m*(a - g) / q

Whether the field is ± is determined by whether the Electrostatic force is attracting or repelling.
 
  • #3
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Got the correct answer! Thanks a lot.
 

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