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)

Question

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)

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 point charge, and r is the distance from the point charge to the point where the potential is being calculated.In this case, the potential at point B is:$$V_B = k\frac{q}{r_B} = (9.0 \times 10^9 \mathrm{~N \cdot m^2/C^2})\frac{(2.0 \times 10^{-6} \mathrm{~C})}{4.0 \mathrm{~m}} = 4500 \mathrm{~V}$$and the potential at point A is:$$V_A = k\frac{q}{r_A} = (9.0 \times 10^9 \mathrm{~N \cdot m^2/C^2})\frac{(2.0 \times 10^{-6} \mathrm{~C})}{6.0 \mathrm{~m}} = 3000 \mathrm{~V}$$Therefore, the potential difference between point B and point A is:$$V_B - V_A = 4500 \mathrm{~V} - 3000 \mathrm{~V} = 1500 \mathrm{~V}$$

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

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