Question

Consider two isolated spherical conductors each having net charge Q. The spheres have radii a and b, where b > a.

Which sphere has the higher potential?
1. the sphere of radius a
2. the sphere of radius b
3. They have the same potential

Answer 1

The sphere with the smaller radius will have a higher potential due to the equation for electric potential, V = kQ/r, where V is the potential, k is Coulomb's constant, Q is the charge, and r is the radius.

Step-by-step explanation:

The electric potential, or voltage, of a conductor depends on the amount of charge it possesses and the size of the conductor. In this case, both spheres have the same net charge, but the sphere with the smaller radius will have a higher potential.

This can be understood through the equation for electric potential: V = kQ/r, where V is the potential, k is Coulomb's constant, Q is the charge, and r is the radius. Since the charge (Q) is the same for both spheres, the smaller sphere (with radius a) will have a smaller radius (r), resulting in a higher potential (V). So, the sphere with radius a has the higher potential.

For example, let's say that both spheres have a net charge of +10C. If the radius of the smaller sphere (a) is 1cm and the radius of the larger sphere (b) is 2cm, the potential for the smaller sphere would be V = (9x10^9 Nm^2/C^2)(10C)/(0.01m) = 9x10^13 V, and the potential for the larger sphere would be V = (9x10^9 Nm^2/C^2)(10C)/(0.02m) = 4.5x10^13 V. Therefore, the sphere of radius a has the higher potential.

Answer 2

1. the sphere of the radius a

Step-by-step explanation:

Because the charge distribution for each case is spherically symmetric, we can choose a spher- ical Gaussian surface of radius r , concentric with the sphere in question.

So E = k (Q /r 2) (for r ≥ R ) , where R is the radius of the sphere being considered, either a or b .

With the choice of potential at r = ∞ being zero, the electric potential at any distance r from the center of the sphere can be expressed as V = - integraldisplay r ∞ E dr = k /Q r

(for r ≥ R ) .

On the spheres of radii a and b , we have V a = k (Q/ a)and V b = k (Q/ b), respectively.

So Since b > a , the sphere of radius a will have the higher potential.

Also recall Because E = 0 inside a conductor, the potential