A sphere with radius 2.0 mm carries a $+ 2.0 \mu \mathrm { C }$ charge.What is the potential difference, $V _ { B } - V _ { A }$ between point B,which is $4.0 \mathrm {~m}$ from the center of the sphere,and point A,which is $6.0 \mathrm {~m}$ from the center of the sphere? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

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Quizplus · Accepted Answer

The potential difference between two points in an electric field is given by the equation:$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{q}{r_B} - \frac{q}{r_A}\right)$$where:* $V_B - V_A$ is the potential difference between points B and A* $q$ is the charge on the sphere* $r_B$ is the distance from point B to the center of the sphere* $r_A$ is the distance from point A to the center of the spherePlugging in the given values, we get:$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{2.0 \mu C}{4.0 \mathrm{~m}} - \frac{2.0 \mu C}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{2.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{2.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left(\frac{1.0 \times 10^{-6} \mathrm{~C}}{4.0 \mathrm{~m}} - \frac{1.0 \times 10^{-6} \mathrm{~C}}{6.0 \mathrm{~m}}\right)$$$$V_B - V_A

A Sphere with Radius 2 +2.0μC+ 2.0 \mu \mathrm { C }+2.0μC

A Sphere with Radius 2 $+ 2.0 \mu \mathrm { C }$