A sphere with radius 2.0 mm carries a 3 μc charge. what is the potential difference, vb - va , between point b 3 m from the center of the sphere and point a 7 m from the center of the sphere? (the value of k is 9.0 × 109 n∙m2/c2.)

5.143 kV

Explanation:

Parameters given:

Radius, r = 2 mm = 0.002 m

Charge, Q = 3 * 10^(-6) C

The potential at any point from the center of a sphere is equivalent to the potential from a point charge.

To find Vb - Va, we have to find the potential at point b, Vb, and the potential at point a, Va.

Potential at point b:

Vb = kQ/b

b = 3m

Vb = (9 * 10^9 * 3 * 10^(-6))/3

Vb = (27 * 10³)/3

Vb = 9 kV

Potential at point a:

Va = kQ/a

a = 7m

Va = (9 * 10^9 * 3 * 10^(-6))/7

Va = (27 * 10³)/7

Va = 3.857 kV

Vb - Va = 9 - 3.857

Vb - Va = 5.143 kV

found in the nucleus

has mass of one amu

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