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lora16 [44]
3 years ago
14

A body that offers little resistance to the movement of free electrons from one point to another is a/an

Physics
2 answers:
stealth61 [152]3 years ago
5 0

Answer: tnks

Explanation: :)))))))

Oksana_A [137]3 years ago
3 0
It should be an Electrical conductor, since conductors allow for movement of electrons  from one point to another. 
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Two tiny particles having charges 20.0 μC and 8.00 μC are separated by a distance of 20.0 cm What are the magnitude and directio
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Answer:

The magnitude and direction of electric field midway between these two charges is 10.8\times10^{5}\ N/C along AB.

Explanation:

Given that,

First charge q_{1}= 20\mu C

second charge q_{2}= 8\mu C

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We need to calculate the electric field

For first charge,

Using formula of electric field

E_{1}= \dfrac{kq_{1}}{r^2}

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E_{1}=\dfrac{9\times10^{9}\times20\times10^{-6}}{10\times10^{-2}}

E_{1}=18\times10^{5}\ N/C

Direction of electric field along AB

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For second charge,

Using formula of electric field

E_{2}= \dfrac{kq_{2}}{r^2}

Put the valueinto the formula

E_{2}=\dfrac{9\times10^{9}\times8\times10^{-6}}{10\times10^{-2}}

E_{2}=7.2\times10^{5}\ N/C

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E_{net}=E_{1}-E_{2}

E_{net}=(18-7.2)\times10^{5}\ N/C

E_{net}=10.8\times10^{5}\ N/C

Direction of net electric field along AB

Hence, The magnitude and direction of electric field midway between these two charges is 10.8\times10^{5}\ N/C along AB.

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