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qwelly [4]
3 years ago
13

Two conducting spheres are each given a charge Q. The radius of the larger sphere is three times greater than that of the smalle

r sphere. If the electric field just outside of the smaller sphere is E0, then the electric field just outside of the larger sphere is
3 Two conducting spheres are each given a charge Q.
1/9 Two conducting spheres are each given a charge Q.
9 Two conducting spheres are each given a charge Q.
1/3 Two conducting spheres are each given a charge Q.
Physics
1 answer:
Vinil7 [7]3 years ago
7 0

Answer:

1/9 of that just outside the smaller sphere

Explanation:

The electric field strength produced by a charged sphere outside the sphere itself is equal to that produced by a single point charge:

E=k\frac{Q}{r^2}

where

k is the Coulomb's constant

Q is the charge on the sphere

r is the distance from the centre of the sphere

Calling R the radius of the first sphere, the electric field just outide the surface of the first sphere is

E_0=k\frac{Q}{R^2}

The second sphere has a radius which is 3 times that of the smaller sphere:

R'=3R

So, the electric field just outside the second sphere is

E'=k\frac{Q}{R'^2}=k\frac{Q}{(3R)^2}=\frac{1}{9}(k\frac{Q}{R^2})=\frac{E_0}{9}

So, the correct answer is 1/9.

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The following is not true about the Law of Conservation of Energy:

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The energy stored in the capacitor quadruples its original value.

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Instead, in this problem the voltage applied is doubled:

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One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other wor
zhannawk [14.2K]

Answer:

Part a)

v = 16.52 m/s

Part b)

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Explanation:

Part a)

(a) when the large-mass object is the one moving initially

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so here we can say

m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v

since this is a perfect inelastic collision so after collision both balls will move together with same speed

so here we can say

v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}

v = \frac{(8.4\times 24 + 3.8\times 0)}{3.8 + 8.4}

v = 16.52 m/s

Part b)

(b) when the small-mass object is the one moving initially

here also we can use momentum conservation as the net force on the system of two masses will be zero

so here we can say

m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v

Again this is a perfect inelastic collision so after collision both balls will move together with same speed

so here we can say

v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}

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A

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