The coach has not seen improvement as expected in his team and plans to change practice session. To help the players he changes
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The question is incomplete. Here is the complete question.
A uniform electric field of 2kN/C points in the +x-direction.
(a) What is the change in potential energy of a +2.00nC test charge, as it is moved from point a at x = -30.0 cm to point b at x = +50.0 cm?
(b) The same test charge is released from rest at point a. What is the kinetic energy when it passes through point b?
(c) If a negative charge instead of a positive charge were used in this problem, qualitatively, how would your answers change?
Answer: (a) ΔU = 3.2× J
(b) KE = 2× J
Explanation: <u>Potential</u> <u>Energy</u> (U) is the amount of work done due to its position or condition and its unit is Joule (J). <u>Kinetic</u> <u>Energy</u> (KE) is the ability to do work by virtue of velocity and the unit is also (J). <u>Mechanical</u> <u>Energy</u> is the sum of Potential and Kinetic Energies of a system.
(a) Related to electricity, Potential Energy can be calculated as:
ΔU = Eqd
where E is the electric field (in N/C);
q is the charge (in C);
d is the distance between plaques (in m);
For a at x = - 30cm and b at x = 50 cm:
E = 2× N/C
q = 2× C
d = 50 - (-30) = 80× = 8×
m
ΔU = = Eqd
= 2×
. 2×
. 8×
ΔU = 3.2× J
(b) Mechanical Energy is constant, so:
Since the initial position is zero and there is no initial kinetic energy:
- (2×
. 2×
. 5×
)
J
(c) If the charge is negative, electric field does positive work, which diminishes the potential energy. The charge flows from the negative side towards the positive side and stays, not doing anything.