98. In Fig. 24-71, a metal sphere with charge q = 5.00 uC and radius

r = 3.00 cm is concentric with a

larger metal sphere with charge Q =

15.0 uC and radius R = 6.00 cm. (a)

What is the potential difference be-

tween the spheres? If we connect the

spheres with a wire, what then is the

charge on (b) the smaller sphere and

(c) the larger sphere?

(a) The potential difference between the spheres is 750 kVA

(b) The charge on the smaller sphere is 6. μC

(c) The charge on the smaller sphere, Q₁ = 13. μC

Explanation:

(a) The given parameters are;

The charge on the inner sphere, q = 5.00 μC

The radius of the inner sphere, r = 3.00 cm = 0.03 m

The charge on the larger sphere, Q = 15.0 μμC

The radius of the larger sphere, R = 6.00 cm = 0.06 m

The potential difference between two concentric spheres is given according to the following equation;

Where;

R = The radius of the larger sphere = 0.06 m

r = The radius of the inner sphere = 0.03 m

q = The charge of the inner sphere = 5.00 × 10⁻⁶ C

Q = The charge of the outer sphere = 15.00 × 10⁻⁶ C

k = 9 × 10⁹ N·m²/C²

Therefore, by plugging in the value of the variables, we have;

The potential difference between the spheres, = 750,000 N·m/C = 750 kVA

(b) When the spheres are connected with a wire, the charge, 'q', on the smaller sphere will be added to the charge, 'Q', on the larger sphere which as follows;

= Q + q = (5 + 15) × 10⁻⁶ C = 20 × 10⁻⁶ C

= 20 × 10⁻⁶ C

From which we have;

Q₁/Q₂ = R/r

Where;

Q₁ = The new charge on the on the larger sphere

Q₂ = The new charge on the on the smaller sphere

= 20 × 10⁻⁶ C = Q₁ + Q₂

∴ Q₁ = 20 × 10⁻⁶ C - Q₂ = 20 μC - Q₂

∴ (20 μC - Q₂)/Q₂ = 0.06/(0.03) = 2

20 μC - Q₂ = 2·Q₂

20 μC = 3·Q₂

Q₂ = 20 μC/3

The charge on the smaller sphere, Q₂ = 20 μC/3 = 6. μC

(c) Q₁ = 20 μC - Q₂ = 20 μC - 20 μC/3 = 40 μC/3

The charge on the smaller sphere, Q₁ = 40 μC/3 = 13. μC.

6 cos 34, 6 sin 34 trig

f=ma newton's second law 7x8=56n