A sphere with radius $2.0 \mathrm {~mm}$ carries a $+ 2.0 \mu \mathrm { C }$ charge. What is the potential difference, $V _ { B } - V _ { A }$, between point $\mathrm { B }$, which is $4.0 \mathrm {~m}$ from the center of the sphere, and point $\mathrm { A }$, which is $6.0 \mathrm {~m}$ from the center of the sphere? $\left( k = 1 / 4 \pi \varepsilon _ { 0 } = 9.0 \times 10 ^ { 9 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 } \right)$

Question

Quizplus · Accepted Answer

The potential difference $V_B - V_A$ between two points outside a charged sphere can be calculated using the formula for the electric potential $V = k \frac{Q}{r}$, where $k = 9.0 \times 10^9 \, \mathrm{N\cdot m^2/C^2}$, $Q$ is the charge, and $r$ is the distance from the center of the charge. For point B, $r_B = 4.0 \, \mathrm{m}$, and for point A, $r_A = 6.0 \, \mathrm{m}$. The potential difference is $V_B - V_A = kQ\left(\frac{1}{r_B} - \frac{1}{r_A}\right)$. Substituting the given values, $V_B - V_A = (9.0 \times 10^9) \times (2.0 \times 10^{-6})\left(\frac{1}{4} - \frac{1}{6}\right) = 1500 \, \mathrm{V}$.

A Sphere with Radius 2.0 mm2.0 \mathrm {~mm}2.0 mm Carries A +2.0μC+ 2.0 \mu \mathrm { C }+2.0μC

A Sphere with Radius $2.0 \mathrm {~mm}$ Carries A $+ 2.0 \mu \mathrm { C }$