A water drop of mass 11.0 mg and having a charge of 1.6 × 10-6 C st

1.	A water drop of mass 11.0 mg and having a charge of 1.6 × 10-6 C stays suspended in a room. What will be the magnitude and direction of electric Held in the room?
Answer» As the drop is suspended, Force (F) due to electric field balances the weight of the drop. ∴ F = mg ………….. (1) Here, m = 11.0 mg = 11 × 10^-6 kg, q = 1.6 × 10^-6 C Electric field is given by, E = \(\frac{F}{q}\) = \(\frac{mg}{q}\) = \(\frac{11\times10^{-6}\times9.8}{1.6\times10^{-6}}\) E = 67.4 N/C As upward force balances the weight, hence direction of electric field must be vertically upwards.

A water drop of mass 11.0 mg and having a charge of 1.6 × 10-6 C stays suspended in a room. What will be the magnitude and direction of electric Held in the room?

Answer»

As the drop is suspended,

Force (F) due to electric field balances the weight of the drop.

∴ F = mg ………….. (1)

Here, m = 11.0 mg

= 11 × 10^-6 kg,

q = 1.6 × 10^-6 C

Electric field is given by,

E = \(\frac{F}{q}\)

= \(\frac{mg}{q}\)

= \(\frac{11\times10^{-6}\times9.8}{1.6\times10^{-6}}\)

E = 67.4 N/C

As upward force balances the weight, hence direction of electric field must be vertically upwards.

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A water drop of mass 11.0 mg and having a charge of 1.6 × 10-6 C stays suspended in a room. What will be the magnitude and direction of electric Held in the room?

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