A 6.9 μC 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 = 1/4πε₀ = 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 charge. The potential difference is given by:$$V = k\frac{q}{r}$$where:* k is Coulomb's constant* q is the charge of the negative object* r is the distance between the objectsPlugging in the values given in the problem, we get:$$V = (9.0 	imes 10^9 \frac{N \cdot m^2}{C^2})\frac{(-6.9 \mu C)}{0.016 m}$$$$V = -3.9 MV$$The negative sign indicates that the potential difference is negative, which means that the positive object is moving from a region of higher potential to a region of lower potential.

A 69 μC Negative Point Charge Has a Positively Charged Particle