Question 2 of 12 Map Sapling Learning macmillan learring A particle that carries

Question

Question 2 of 12 Map Sapling Learning macmillan learring A particle that carries a net charge of-41.8 C is held in a region of constant, uniform electric field. The electric field vector is oriented 40.2° clockwise from the vertical axis, as shown. If the magnitude of the electric field is 5.82 N/C, how much work is done by the electric field as the particle is made to move a distance of d = 0.356 m straight up? Number 40.2 What is the potential difference between the particle's initial and final positions (V-V)? Number

Explanation / Answer

Given: q = -41.8 µC = -41.8 * 10-6 C, E = 5.82 N/C, d = 0.356 m, = 40.20

Solution: 1) Along the vertical direction electric field component E = E*cos = 5.82 * cos 40.20

i.e. E = 4.445 N/C

We know that,

Work done W = F*d = q*E*d = 41.8 * 10-6 * 4.445 * 0.356 = 66.15 * 10-6 J

2) To find the potential difference Vf – Vi between final and initial position of the particle when it is displaced through d = 0.356 m in the electric field of 5.82 N/C is given by,

V = (E * cos) * d = 4.445 * 0.356 = 1.58 volt (considering magnitude only)

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