A particle that carries a net charge of -41.8 μc is held in a region of constant, uniform electric field. the electric field vector is oriented 55.2° clockwise from the vertical axis, as shown. if the magnitude of the electric field is 4.82 n/c, how much work is done by the electric field as the particle is made to move a distance of d = 0.556 m straight up?

The total work done by the electric field on the charge is given by the scalar product between the electric force acting on the charge and the displacement of the charge:
[tex]W=F d cos \theta[/tex]
where the force is F=qE, d=0.556 and [tex]\theta=55.2^{\circ}[/tex]. Using the value of q and E given by the problem, we find
[tex]W=qEdcos\theta = 6.39\cdot10^{-5}J[/tex]

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priyankatutor priyankatutor · Answer 2 · 2022-02-04T13:35:52+01:00

The work done by the electric field as the particle is made to move a given distance is [tex]W = 6.39\times 10^{-5} \rm \ J[/tex].

How to calculate work done by the electric field on charge?

The work can be calculated by multiplying the force exerted on the charge and the displacement of the charge.

Since the charge is moving in the verticle direction,

[tex]W = Fdcos \theta[/tex]

Where,

[tex]F[/tex] - force = qE =-41.8 μc x 4.82 N/C = -201.47

[tex]d[/tex] - displacement = 0.556 m

[tex]\theta[/tex] - angle = 55.2°

Put the values in the formula,

[tex]W = -201.47 \times 0.556 {\rm \ m \times cos 55.2)}\\\\W = 6.39\times 10^{-5} \rm \ J[/tex]

Therefore, the work done by the electric field as the particle is made to move a given distance is [tex]W = 6.39\times 10^{-5} \rm \ J[/tex].

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