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makvit [3.9K]
3 years ago
12

A thin spherical shell with radius R1 = 2.00 cm is concentric with a larger thin spherical shell with radius R2 = 6.00 cm. Both

shells are made of insulating material. The smaller shell has charge q1=+6.00nC distributed uniformly over its surface, and the larger shell has charge q2=−9.00nC distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells.
(a) What is the electric potential due to the two shells at the following distance from their common center: (i) r=0; (ii) r=4.00cm; r=6.00cm?

(b) What is the magnitude of the potential difference between the surfaces of the two shells? Which shell is at the higher potential: the inner shell or the outer shell?
Physics
1 answer:
Dafna11 [192]3 years ago
8 0

Answer:

a. i. 1350 V ii 0 V iii -450 V b. 6.75 kV. The inner shell is at a higher potential.

Explanation:

The formula for electric potential is given by V = Σkq/r, where k = 9 × 10⁹ Nm²/C², q = charge and r = distance.

q₁ = charge on smaller shell = +6.00 nC = +6.00 × 10⁻⁹ C, r₁ = radius of smaller shell = 2.00 cm = 2.00 × 10⁻² m.

q₂ = charge on larger shell = -9.00 nC = -9.00 × 10⁻⁹ C, r₂ = radius of larger shell = 6.00 cm = 6.00 × 10⁻² m.

a. At r = 0, inside both spheres V = kq₁/r₁ + kq₂/r₂. = k(q₁/r₁ + q₂/r₂) = 9 × 10⁹ [+6.00 × 10⁻⁹/2.00 × 10⁻² + (-9.00 × 10⁻⁹/6.00 × 10⁻²)] = 1350 V

ii. At r = 4.00 cm, the point outside of smaller shell but inside larger shell. r₁ = 4.00 cm = 4.00 × 10⁻² m and r₂ = 6.00 cm = 6.00 × 10⁻². So, V = kq₁/r₁ + kq₂/r₂. = k(q₁/r₁ + q₂/r₂) = 9 × 10⁹ [+6.00 × 10⁻⁹/4.00 × 10⁻² + (-9.00 × 10⁻⁹/6.00 × 10⁻²)] = 0 V.

iii. At r = 6.00 cm, the point outside both shells. r₁ = r₂ = r = 6.00 cm = 6.00 × 10⁻². So, V = kq₁/r₁ + kq₂/r₂. = k(q₁ + q₂)/r = 9 × 10⁹ [+6.00 × 10⁻⁹+ (-9.00 × 10⁻⁹)]/6.00 × 10⁻² = -450 V.

b. The potential of the surface of the smaller shell is V₁ = 9 × 10⁹ [+6.00 × 10⁻⁹/2.00 × 10⁻²] = 2700 V = 2.7 kV.

The potential of the surface of the larger shell is V₂ = 9 × 10⁹ [-9.00 × 10⁻⁹/2.00 × 10⁻²] = -4050 V = -4.050 kV. The potential difference V₁ - V₂ = 2700 - (-4050) V = 6750 V = 6.75 kV. Since the potential difference is positive, V₁ is higher. So, the inner shell is at a higher potential.

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How large a force is necessary to stretch a 4.0-mm-diameter steel wire from its original length by 1.0%?
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The force needed to stretch the steel wire by 1% is 25,140 N.

The given parameters include;

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  • final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)
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The area of the steel wire is calculated as follows;

A = \pi r^2\\\\ A= 3.142 \times (0.002)^2\\\\ A= 1.257 \times 10^{-5} \ m^2

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E = \frac{stress}{strain} = \frac{F/A}{e/L} = \frac{FL}{Ae} \\\\F = \frac{EAe}{L}

F = \frac{200 \times 10^9\  \times\  1.257\times 10^{-5}\  \times \ 0.01l_1}{l_1} \\\\F = 25,140\ N

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Learn more here: brainly.com/question/21413915

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E_{cl} = longitudinal modulus of elasticity = 35 Gpa

E_{ct} = transverse modulus of elasticity = 5.17 Gpa

E_m = Epoxy modulus of elasticity = 3.4 Gpa

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V_{\rho t} = Volume fraction of fibre (transvers)

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E_{cl}=E_m(1-V_{\rho l})+E_fV_{\rho l}\\\Rightarrow 35=3.4(1-V_{\rho l})+131V_{\rho l}\\\Rightarrow 35=3.4-3.4V_{\rho l}+131V_{\rho l}\\\Rightarrow V_{\rho l}=\frac{35-3.4}{131-3.4}\\\Rightarrow V_{\rho l}=0.24764

Transverse modulus of elasticity is given by

E_{ct}=\frac{E_mE_f}{(1-V_{\rho t})E_f+V_{\rho t}E_m}\\\Rightarrow 5.17=\frac{3.4\times 131}{(1-V_{\rho t})131+V_{\rho t}3.4}\\\Rightarrow \frac{3.4\times 131}{5.17}-131=-127.6V_{\rho t}\\\Rightarrow V_{\rho t}=\frac{\frac{3.4\times 131}{5.17}-131}{-127.6}\\\Rightarrow V_{\rho t}=0.35148

V_{\rho l}\neq V_{\rho t}

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