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Nadusha1986 [10]
3 years ago
10

An electron in a vacuum chamber is fired with a speed of 9800 km/s toward a large, uniformly charged plate 75 cm away. The elect

ron reaches a closest distance of 15 cm before being repelled.
Physics
1 answer:
melisa1 [442]3 years ago
3 0

Answer:

The plate's surface charge density is -8.056\times10^{-9}\ C/m^2

Explanation:

Given that,

Speed = 9800 km/s

Distance d= 75 cm

Distance d' =15 cm

Suppose we determine the plate's surface charge density?

We need to calculate the surface charge density

Using work energy theorem

W=\Delta K.E

W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{i}^2

Here, final velocity is zero

W=0-\dfrac{1}{2}mv_{i}^2...(I)

We know that,

W=-Fd

W=-E\times e\times d

W=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d...(II)

From equation (I) and (II)

-\dfrac{1}{2}mv_{i}^2=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d

Charge is negative for electron

\lambda=\dfrac{mv^2\epsilon_{0}}{(-e)d}

Put the value into the formula

\lambda=-\dfrac{9.1\times10^{-31}\times(9800\times10^{3})^2\times8.85\times10^{-12}}{1.6\times10^{-19}\times(75-15)\times10^{-2}}

\lambda=-8.056\times10^{-9}\ C/m^2

Hence, The plate's surface charge density is -8.056\times10^{-9}\ C/m^2

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