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How to Visualize Electric Fields

A charged oil...

Question

A charged oil drop is suspended in a uniform field of 3 x 10⁴ v/m so that it neither falls nor rises. The charge on the drop will (Take the mass of the charge = 9.9 x 10^-15 kg and g = 10 m/s² )

3.3 \times 10^{- 18} N / C

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2.25 \times 10^{10} N / C

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5 \times 10^{10} N / C

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None

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Solution

The correct option is A $3.3 \times 10^{- 18} N / C$
$\begin{matrix} A t e q u i l i b r i u m, e l e c t r i c f o r c e o f d r o p b a l a n c e s w e i g h t o f d r o p . q E = m g q = \frac{m g}{E} N o w, q = \frac{9.9 \times 10^{- 15}}{3 \times 10^{4}} = 3.3 \times 10^{- 18} C \end{matrix}$

Suggest Corrections

Similar questions

3 \times 10^{4} V / m

= 9.9 \times 10^{- 15}

g = 10 m / s^{2}

3 \times 10^{4} V / m

= 9.9 \times 10^{- 15} a n d g = 10

3.2 \times 10^{- 18}

1.6 \times 10^{- 19}

g = 10

m / s^{2}

1.6 \times 10^{- 15} N

2 \times 10^{3} N C^{- 1}

2.5 \times 10^{- 7} k g

2 \times 10^{- 2} m^{2}

5 \times 10^{- 7} C

g = 10 {m / s}^{2}

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