The electric field that can balance a charged particle of mass 3-2- 10 -27 kg is Given that the charge on the particle is 1-6- 10 -19 C then

The correct option is A 19.6× 10 ^ 8 N C ^ 1 here m=3.2× 10 ^ 27 kg,e=1.6× 10 ^ 19 C,g=9.8 ms ^ 2 As #10; E= F q = mg e = 3.2× ...

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Standard XII

Physics

Force on a Current Element

The electric ...

Question

The electric field that can balance a charged particle of mass $3.2 \times 10^{- 27} k g$ is (Given that the charge on the particle is $1.6 \times 10^{- 19} C$ ) then

19.6 \times 10^{- 8} N C^{- 1}

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20 \times 10^{- 6} N C^{- 1}

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19.6 {\times 10}^{8} N C^{- 1}

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20 \times 10^{6} N C^{- 1}

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Solution

The correct option is A $19.6 \times 10^{- 8} N C^{- 1}$
here $m = 3.2 \times 10^{- 27} k g, e = 1.6 \times 10^{- 19} C, g = 9.8 {m s}^{- 2}$
As $E = \frac{F}{q} = \frac{m g}{e}$
$= \frac{3.2 \times 10^{- 27} \times 9.8}{1.6 \times 10^{- 19}}$ $= 19.6 \times 10^{- 8} N C^{- 1}$

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The electric field that can balance a charged particle of mass 3.2×10−27kg is (Given that the charge on the particle is 1.6×10−19C) then

The correct option is A 19.6×10−8NC−1here m=3.2×10−27kg,e=1.6×10−19C,g=9.8ms−2As E=Fq=mge=3.2×10−27×9.81.6×10−19 =19.6×10−8NC−1

The electric field that can balance a charged particle of mass $3.2 \times 10^{- 27} k g$ is (Given that the charge on the particle is $1.6 \times 10^{- 19} C$ ) then

The correct option is A $19.6 \times 10^{- 8} N C^{- 1}$
here $m = 3.2 \times 10^{- 27} k g, e = 1.6 \times 10^{- 19} C, g = 9.8 {m s}^{- 2}$
As $E = \frac{F}{q} = \frac{m g}{e}$
$= \frac{3.2 \times 10^{- 27} \times 9.8}{1.6 \times 10^{- 19}}$ $= 19.6 \times 10^{- 8} N C^{- 1}$