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?

6.4×10⁻¹⁹ C

Explanation:

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.

Also Using,

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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