Sapling Learning A particle that carries a net charge of -59.8 UC is held in a r

Question

Sapling Learning A particle that carries a net charge of -59.8 UC is held in a region of constant, uniform electric field. The electric field vector is oriented 25.2 clockwise from the vertical axis, as shown. If the magnitude of the electric field is 6.82 N/C, how much work by fieldas the particle is made to move a distance of d 0.156 m straight up? Incorrect. The angle involved n the work calculation should d represent the angle between Number the force and displacement u ectors. The force vector does 57.567 x 10 not poin in the same direction as the field. What is the potential difference between the particle's initial and final positions (Vi-V)? Number .96266 Previous Give Up & View solution a Try Again Next

Explanation / Answer

Electric field along the vertical direction is E*cos(25.2) = 6.82*cos(25.2) = 6.17 N/C

Force along vertical direction is F = q*E*cos(theta) = 59.8*10^-6*6.17 = 3.68*10^-4 N along vertically downwards

Work done is W = -F*S = -3.68*10^-4*0.156 = -5.74*10^-5 J = -57.4*10^-6 J

Potential difference is Vf-Vi = -E*cos(theta)*d = -6.17*0.156 = -0.96252 V

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