A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/m

Part A What work is done by the electric force when the charge moves a distance of 0.480 m to the right?

Part B What work is done by the electric force when the charge moves a distance of 0.660 m upward?

Part C What work is done by the electric force when the charge moves a distance of 2.50 m at an angle of 45.0° downward from the horizontal?

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

[tex]W=Fd cos \theta[/tex]

where

F is the magnitude of the force

d is the displacement of the particle

[tex]\theta[/tex] 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

[tex]\theta=90^{\circ}[/tex]

and [tex]cos 90^{\circ}=0[/tex]: 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

[tex]\theta=0^{\circ}[/tex]

and

[tex]cos 0^{\circ}=1[/tex]

Therefore, the work done by the electric force is

[tex]W=Fd[/tex]

and we have:

[tex]F=qE[/tex] is the magnitude of the electric force. Since

[tex]E=4.30\cdot 10^4 V/m[/tex] is the magnitude of the electric field

[tex]q=32.0 nC = 32.0\cdot 10^{-9}C[/tex] is the charge

The electric force is

[tex]F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N[/tex]

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

[tex]W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J[/tex]

C)

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

[tex]\theta=90^{\circ}+45^{\circ}=135^{\circ}[/tex]

Moreover, we have:

[tex]F=1.38\cdot 10^{-3} N[/tex] (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:

[tex]W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J[/tex]

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:

brainly.com/question/6763771

brainly.com/question/6443626

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly