Determine The Speed of a Charge at a given point

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The discussion revolves around calculating the speed of a charged particle using the electric potential function V= 3x^2 + 2xy + (y^2)z + 20. The participant attempts to find the speed of a 4 coulombs, 2kg charge released from the origin at the point (1,1,-1) by equating the change in potential energy to the change in kinetic energy. They calculate a change in electric potential of 4 V, leading to a negative value when solving for velocity, indicating a potential error in the problem setup. The consensus is that the question is flawed, as a positive charge cannot move from a lower to a higher potential without external force. The discussion highlights the importance of correctly interpreting electric potential in relation to charge movement.
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Homework Statement


Given: The electric potential function: V= 3x^2 + 2xy +(y^2)z + 20
a) A 4 coulombs, 2kg charge is released from the origin. Determine the speed of the charge at (1,1,-1)

Homework Equations


Kinetic Energy= 1/2 mv^2

The Attempt at a Solution


I tried to use the fact that the change in potential energy is equal to the negative change in kinetic energy.
I calculated the change in electric potential to be 4 V and set that equal to the negative change in kinetic energy, which would just be -1/2 mv^2
However, when I solve for the velocity, I get the square root of negative 4. Have I done something wrong or is the problem just flawed?
 
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I think you're right. The question is flawed. A positive charge will not go from lower to higher potential without some external force.
 

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