A -7.0-μC point charge has a positively charged object in an elliptical orbit around it.If the mass of the positively charged object is 1.0 kg and the distance varies from 5.0 mm to 20.0 mm between the charges,what is the maximum electric potential difference through which the positive object moves? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²) A)9.4 MV B)3.2 MV C)4.2 MV D)16 MV

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Quizplus · Accepted Answer

The maximum electric potential difference is calculated using the formula for electric potential $V = \frac{kQ}{r}$, where $k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$, $Q = -7.0 \, \mu\text{C} = -7.0 \times 10^{-6} \, \text{C}$, and $r$ is the distance between the charges. The maximum potential difference occurs between the closest approach (5.0 mm = 0.005 m) and the farthest distance (20.0 mm = 0.020 m). Thus, the potential difference $\Delta V = V_{\text{far}} - V_{\text{close}} = \frac{kQ}{r_{\text{far}}} - \frac{kQ}{r_{\text{close}}}$. Substituting the given values, $\Delta V = \frac{8.99 \times 10^9 \times -7.0 \times 10^{-6}}{0.020} - \frac{8.99 \times 10^9 \times -7.0 \times 10^{-6}}{0.005}$, results in a maximum electric potential difference of approximately 9.4 MV.

A -70-μC Point Charge Has a Positively Charged Object in an an Elliptical