28 August, 14:23
As a particle moves 10.0 m along an electric field of strength 75 N/C, its electrical potential energy decreases by 4.8 * 10-16 J. What is the particle's charge?
28 August, 15:22
From the question,
V = E*d ... Equation 1
Where V = electric potential, E = Electric Field, d = distance moved by the particle.
Given: E = 75 N/C, d = 10 m.
Substitute into equation 1
V = 75*10 = 750 V.
W = qV ... Equation 2
W = Work done by the particle when it moves, q = Particles charge
make q the subject of the equation,
q = W/V ... Equation 3
Given: W = 4.8*10⁻¹⁶ J, V = 750 V
Substitute into equation 3
q = 4.8*10⁻¹⁶/750
q = 6.4*10⁻¹⁹ C
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» As a particle moves 10.0 m along an electric field of strength 75 N/C, its electrical potential energy decreases by 4.8 * 10-16 J. What is the particle's charge?