Question

A liquid drop having charge – q is placed in a uniform electric field E acting
vertically downwards . Its weight is W if the drop in equilibrium in air , the
electric field E is equal to :
(a) W/q
(b) W x E
(c) W x E²
(d) W/E²

Answer 1

Answer:

Given : Electric field, E=3×10

4

V/m

Mass of the drop, M=9.9×10

−15

kg

Let q be the amount of the charge that the drop carries.

The coulomb force balances the gravitational force acting on the drop at equilibrium.

∴ qE=Mg ⟹q=

E

Mg

∴ q=

3×10

4

9.9×10

−15

×10

=3.3×10

−18

C

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Answer 2

Answer:

Correct option is A

3.3×10

−18

C

Given : Electric field, E=3×10

4

V/m

Mass of the drop, M=9.9×10

−15

kg

Let q be the amount of the charge that the drop carries.

The coulomb force balances the gravitational force acting on the drop at equilibrium.

∴ qE=Mg ⟹q=

E

Mg

∴ q=

3×10

4

9.9×10

−15

×10

=3.3×10

−18

C Electric field, E=3×10

4

V/m

Mass of the drop, M=9.9×10

−15

kg

Let q be the amount of the charge that the drop carries.

The coulomb force balances the gravitational force acting on the drop at equilibrium.

∴ qE=Mg ⟹q=

E

Mg

∴ q=

3×10

4

9.9×10

−15

×10

=3.3×10

−18

C