A sphere with radius 2.0 mm carries +1.0 μC of charge distributed uniformly throughout its volume.What is the potential difference,V_B - V_A,between point B,which is 4.0 m from the center of the sphere,and point A,which is 9.0 m from the center of the sphere? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²) A)1200 V B)-1200 V C)140 V D)-0.45 V

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Quizplus · Accepted Answer

The potential difference between two points in an electric field is given by the negative of the integral of the electric field from one point to the other. In this case, the electric field is due to a point charge, so the potential difference is given by:$$V_B - V_A = -\int_A^B \frac{kq}{r^2} dr$$where k is Coulomb's constant, q is the charge, and r is the distance from the point charge.Plugging in the values given in the problem, we get:$$V_B - V_A = -\int_9^4 \frac{(8.99 	imes 10^9 N \cdot m^2/C^2)(1.0 	imes 10^{-6} C)}{r^2} dr$$$$= -8.99 	imes 10^3 V \left[\frac{1}{4} - \frac{1}{9}ight]$$$$= -8.99 	imes 10^3 V \left(\frac{5}{36}ight)$$$$= -1200 V$$Therefore, the potential difference between point B and point A is -1200 V.

A Sphere with Radius 2