Question

98. In Fig. 24-71, a metal sphere

Answer 1

Answer:

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

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

(c) The charge on the smaller sphere, Q₁ = 13.$\overline 3$ 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;

$V_r - V_R = k \times q \times \left ( \dfrac{1}{r} - \dfrac{1}{R} \right)$

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;

$V_r - V_R = 9 \times 10^9 \times 5.00 \times 10^{-6} \times \left ( \dfrac{1}{0.03} - \dfrac{1}{0.06} \right) = 750,000$

The potential difference between the spheres, $V_r - V_R$ = 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;

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

$Q_f$ = 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

$Q_f$ = 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.$\overline 6$ 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.$\overline 3$ μC.