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coldgirl [10]
3 years ago
6

A charge +1.9 μC is placed at the center of the hollow spherical conductor with the inner radius 3.8 cm and outer radius 5.6 cm.

Suppose the conductor initially has a net charge of +3.8 μC instead of being neutral. What is the total charge (a) on the interior and (b) on the exterior surface?
Physics
1 answer:
Archy [21]3 years ago
5 0

To solve this problem we will apply the concepts related to load balancing. We will begin by defining what charges are acting inside and which charges are placed outside.

PART A)

The charge of the conducting shell is distributed only on its external surface. The point charge induces a negative charge on the inner surface of the conducting shell:

Q_{int}=-Q1=-1.9*10^{-6} C. This is the total charge on the inner surface of the conducting shell.

PART B)

The positive charge (of the same value) on the external surface of the conducting shell is:

Q_{ext}=+Q_1=1.9*10^{-6} C

The driver's net load is distributed through its outer surface. When inducing the new load, the total external load will be given by,

Q_{ext, Total}=Q_2+Q_{ext}

Q_{ext, Total}=1.9+3.8

Q_{ext, Total}=5.7 \mu C

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Answer:

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8 0
2 years ago
2
luda_lava [24]

Answer:

500000N/m²

5250N

Explanation:

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Pressure is the force per unit area on a body. In fluids, pressure is the product of density, gravity and height

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B.

width of window = 150mm

height of window = 70mm

Force water exerts on the window = ?

To solve this problem;

         Pressure  = \frac{Force}{Area}

Area of the window = width x height  = 150 x 10⁻³ x 70 x 10⁻³

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