A) The work done by the electric field is zero
B) The work done by the electric field is ![9.1\cdot 10^{-4} J](https://tex.z-dn.net/?f=9.1%5Ccdot%2010%5E%7B-4%7D%20J)
C) The work done by the electric field is ![-2.4\cdot 10^{-3} J](https://tex.z-dn.net/?f=-2.4%5Ccdot%2010%5E%7B-3%7D%20J)
Explanation:
A)
The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).
The work done by a force is given by the equation
![W=Fd cos \theta](https://tex.z-dn.net/?f=W%3DFd%20cos%20%5Ctheta)
where
F is the magnitude of the force
d is the displacement of the particle
is the angle between the direction of the force and the direction of the displacement
In this problem, we have:
- The force is directed vertically upward (because the field is directed vertically upward)
- The charge moves to the right, so its displacement is to the right
This means that force and displacement are perpendicular to each other, so
![\theta=90^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D90%5E%7B%5Ccirc%7D)
and
: therefore, the work done on the charge by the electric field is zero.
B)
In this case, the charge move upward (same direction as the electric field), so
![\theta=0^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D0%5E%7B%5Ccirc%7D)
and
![cos 0^{\circ}=1](https://tex.z-dn.net/?f=cos%200%5E%7B%5Ccirc%7D%3D1)
Therefore, the work done by the electric force is
![W=Fd](https://tex.z-dn.net/?f=W%3DFd)
and we have:
is the magnitude of the electric force. Since
is the magnitude of the electric field
is the charge
The electric force is
![F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N](https://tex.z-dn.net/?f=F%3D%2832.0%5Ccdot%2010%5E%7B-9%7D%29%284.30%5Ccdot%2010%5E4%29%3D1.38%5Ccdot%2010%5E%7B-3%7D%20N)
The displacement of the particle is
d = 0.660 m
Therefore, the work done is
![W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J](https://tex.z-dn.net/?f=W%3DFd%3D%281.38%5Ccdot%2010%5E%7B-3%7D%29%280.660%29%3D9.1%5Ccdot%2010%5E%7B-4%7D%20J)
C)
In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is
![\theta=90^{\circ}+45^{\circ}=135^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D90%5E%7B%5Ccirc%7D%2B45%5E%7B%5Ccirc%7D%3D135%5E%7B%5Ccirc%7D)
Moreover, we have:
(electric force calculated in part b)
While the displacement of the charge is
d = 2.50 m
Therefore, we can now calculate the work done by the electric force:
![W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J](https://tex.z-dn.net/?f=W%3DFdcos%20%5Ctheta%20%3D%20%281.38%5Ccdot%2010%5E%7B-3%7D%29%282.50%29%28cos%20135.0%5E%7B%5Ccirc%7D%29%3D-2.4%5Ccdot%2010%5E%7B-3%7D%20J)
And the work is negative because the electric force is opposite direction to the displacement of the charge.
Learn more about work and electric force:
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