Respuesta :
Answer:
(a) The potential difference between the spheres is 750 kVA
(b) The charge on the smaller sphere is 6.[tex]\overline 6[/tex] μC
(c) The charge on the smaller sphere, Q₁ = 13.[tex]\overline 3[/tex] 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;
[tex]V_r - V_R = k \times q \times \left ( \dfrac{1}{r} - \dfrac{1}{R} \right)[/tex]
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;
[tex]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[/tex]
The potential difference between the spheres, [tex]V_r - V_R[/tex] = 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;
[tex]Q_f[/tex] = Q + q = (5 + 15) × 10⁻⁶ C = 20 × 10⁻⁶ C
[tex]Q_f[/tex] = 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
[tex]Q_f[/tex] = 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.[tex]\overline 6[/tex] 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.[tex]\overline 3[/tex] μC.
The potential difference is 7.5*10^5V and the charge on the smaller sphere is 6.67uC while the charge on the larger sphere is 13.34uC
Data;
- q = 5.0 uC
- r = 3.0cm= 0.03m
- Q = 15.0uC
- R = 6.0cm = 0.06m
Potential Difference between the Spheres
for the inner sphere;
[tex]v_i = \frac{kq}{r} = \frac{(9*10^9)(5*10^-^6)}{0.03} \\v_i = 1.5*10^6V[/tex]
for the outer sphere;
[tex]v_o = \frac{KQ}{R} = \frac{(9*10^9)(15*10^-^6)}{0.06} = 2.25 * 10^6V[/tex]
The difference in potential is
[tex]\delta V = v_o - v_i = 2.25*10^6 - 1.5*10^6 = 7.5*10^5V[/tex]
Charge on the Smaller Plate
[tex]q + Q = 5 + 15 = 20 \mu C = 20*10^-^6C ..eq(i)[/tex]
The sharing of charge continues till they attain a point of equal potential
[tex]v_i = v_o \\\frac{kq}{r} = \frac{kQ}{R} \\\frac{q}{0.03} = \frac{Q}{0.06} \\Q= 2q ...eq(ii)\\[/tex]
let's solve for equation (i) and equation (ii)
[tex]q+2q = 20\mu C\\q = 6.67 \mu C[/tex]
The charge on the smaller sphere is 6.67uC
The charge on the larger sphere
The charge on the larger sphere is
[tex]Q = 2q\\q = 6.67 \mu C\\Q = 2 * 6.67 = 13.34 \mu C[/tex]
The charge on the larger sphere is 13.34uC
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