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4 years ago

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An electron with a charge value of 1.6x10^-19 C is in the presence of an electric field of 300 N/C when located between a pair o

f oscilloscope deflecting plates. What force does the electron experience?

1 answer:

Blababa [14]4 years ago

6 0

(1.6 x 10⁻¹⁹ C) x (300 N/C) = <em>4.8 x 10⁻¹⁷ Newton</em>

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Answer:

(A) Capacitance per unit length = $4.02 \times 10^{-10}$

(B) The magnitude of charge on both conductor is $Q = 4.22 \times 10^{-19}$ C and the sign of charge on inner conductor is $+Q$ and the sign on outer conductor is $-Q$

Explanation:

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Radius of inner part of conductor $(R_{1})$ = $2.7 \times 10^{-3}$ m

Radius of outer part of conductor $(R_{2})$ = $3.1 \times 10^{-3}$ m

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(A)

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From the formula of cylindrical capacitor,

$C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }$

Where, $\epsilon_{o} = 8.85 \times 10^{-12}$

But we need capacitance per unit length so,

$\frac{C}{l} = \frac{2\pi\epsilon_{o} }{ln\frac{R_{2} }{R_{1} } }$

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(B)

The charge on both conductors is given by,

$Q = C \Delta V$

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The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is $+Q$ and Charge on outer conductor is $-Q$ .

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