he figure below shows a charged particle, with a charge of q = +38.0 nC, that moves a distance of d = 0.185 m from point A to point B in the presence of a uniform electric field  E  of magnitude 245 N/C, pointing right. A positive point charge q is initially at point A, then moves a distance d to the right to point B. Electric field vector E points to the right. (a) What is the magnitude (in N) and direction of the electric force on the particle? magnitude Ndirection ---Select--- toward the right toward the left The magnitude is zero. (b) What is the work (in J) done on the particle by the electric force as it moves from A to B?  J (c) What is the change of the electric potential energy (in J) as the particle moves from A to B? (The system consists of the particle and all its surroundings.) PEB − PEA =  J

Question
The figure below shows a charged particle, with a charge of q = +38.0 nC, that moves a distance of d = 0.185 m from point A to point B in the presence of a uniform electric field 
E
 of magnitude 245 N/C, pointing right.
A positive point charge q is initially at point A, then moves a distance d to the right to point B. Electric field vector E points to the right.
(a)
What is the magnitude (in N) and direction of the electric force on the particle?
magnitude Ndirection ---Select--- toward the right toward the left The magnitude is zero.
(b)
What is the work (in J) done on the particle by the electric force as it moves from A to B?
 J
(c)
What is the change of the electric potential energy (in J) as the particle moves from A to B? (The system consists of the particle and all its surroundings.)
PEB − PEA =  J
(d)
What is the potential difference (in V) between A and B?
VB − VA =  V
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