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
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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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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