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 = 1/4??₀ = 9.0 × 10⁹ N ? m²/C²)

Question

Quizplus · Accepted Answer

The potential difference between two points in an electric field is given by the negative of the work done per unit charge in moving a charge from one point to the other. In this case, the work done in moving a charge from point A to point B is given by:$$W = q(V_B - V_A)$$where q is the charge of the sphere, and $V_B$ and $V_A$ are the potentials at points B and A, respectively. The potential at a point due to a point charge is given by:$$V = k\frac{q}{r}$$where k is Coulomb's constant, q is the charge of the sphere, and r is the distance from the point to the center of the sphere. Substituting these expressions into the equation for the work done, we get:$$W = q\left(k\frac{q}{r_B} - k\frac{q}{r_A}\right)$$$$W = k\frac{q^2}{r_B} - k\frac{q^2}{r_A}$$$$W = k\frac{q^2}{r_A}\left(\frac{r_A}{r_B} - 1\right)$$$$W = k\frac{q^2}{r_A}\left(\frac{6.0 \mathrm{~m}}{4.0 \mathrm{~m}} - 1\right)$$$$W = k\frac{q^2}{r_A}\left(\frac{2}{4} - 1\right)$$$$W = k\frac{q^2}{r_A}\left(-\frac{1}{2}\right)$$$$W = -\frac{1}{2}k\frac{q^2}{r_A}$$The potential difference between points B and A is then given by:$$V_B - V_A = -\frac{W}{q}$$$$V_B - V_A = -\left(-\frac{1}{2}k\frac{q^2}{r_A}\right)$$$$V_B - V_A = \frac{1}{2}k\frac{q^2}{r_A}$$$$V_B - V_A = \frac{1}{2}\left(9.0 \times 10^9 \mathrm{~N \cdot m^2/C^2}\right)\frac{(2.0 \times 10^{-6} \mathrm{~C})^2}{(6.0 \mathrm{~m})}$$$$V_B - V_A = 1500 \mathrm{~V}$$Therefore, the potential difference between point B and point A is 1500 V.

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

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