A point charge q = +41.0 µC moves from A to B separated by a distance d = 0.178
ID: 2044300 • Letter: A
Question
A point charge q = +41.0 µC moves from A to B separated by a distance d = 0.178 m in the presence of an external electric field of magnitude 289 N/C directed toward the right as in the following figure.
A+-----------d-------------B+
q
(a) Find the electric force exerted on the charge.
magnitude ____________N
direction________
---Select--- 1.toward the right 2. toward the left3. The magnitude is zero.
(b) Find the work done by the electric force.
_____________J
(c) Find the change in the electric potential energy of the charge.
_____________J
(d) Find the potential difference between A and B.
VB - VA = V ______________
Explanation / Answer
Given that q = 41.0 x 10^-6 C distance d =0.178 m electric field E = 289 N/Ca) The electric force exerted on the charge is F = Eq = (289 N/C)( 41.0 x 10^-6 C) =0.011849 N b) The work done by the electric force is W = Fd = (0.011849 N)(0.178 m) = 2.109 x10^-3 J c) The change in the electric potential energy of the charge is V = W/q = (2.109 x10^-3 J)/(41.0 x 10^-6 C) = 51.442 V Given that q = 41.0 x 10^-6 C distance d =0.178 m electric field E = 289 N/C
a) The electric force exerted on the charge is F = Eq = (289 N/C)( 41.0 x 10^-6 C) =0.011849 N b) The work done by the electric force is W = Fd = (0.011849 N)(0.178 m) = 2.109 x10^-3 J c) The change in the electric potential energy of the charge is V = W/q = (2.109 x10^-3 J)/(41.0 x 10^-6 C) = 51.442 V c) The change in the electric potential energy of the charge is V = W/q = (2.109 x10^-3 J)/(41.0 x 10^-6 C) = 51.442 V
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