A 6.9 ?C negative point charge has a positively charged particle in an elliptical orbit about it. If the mass of the positively charged particle is $1.0 \mu \mathrm { g }$ and its distance from the point charge varies from $4.0 \mathrm {~mm}$ to $16.0 \mathrm {~mm}$ , what is the maximum potential difference through which the positive object moves? (k = 1/4??₀ = 9.0 × 10⁹ N ? m²/C²)

Question

Quizplus · Accepted Answer

The maximum potential difference is when the positive object is at its farthest point from the negative point charge. At this point, the potential difference is:$$V = k\frac{q}{r} = (9.0 	imes 10^9 	ext{ N}\cdot	ext{m}^2/	ext{C}^2)\frac{(-6.9 	imes 10^{-6} 	ext{ C})}{0.016 	ext{ m}} = -12 	ext{ MV}$$The negative sign indicates that the potential difference is in the opposite direction of the electric field.

A 69 ?C Negative Point Charge Has a Positively Charged Particle