If the strength of the earth's electric field is 100 N/CN/C , how does the magnitude of the electric force on the droplet compare to the weight force

Respuesta :

Magnitude of electric force = 15 x 10⁻¹⁰ N

Magnitude of weight force = 64.092 x 10⁻⁶ N

Given

Diameter of rain drop = 2.5mm

Radius = 1.25 mm = 1.25 × 10⁻³ m/sec

Drop has a charge = +15pC

15pC = 15×10⁻¹²C

Electric field (E) = 100 N/C

Electric force on the charge

= charge x electric field = qE

= 15 x 10⁻¹² x 100

= 15 x 10⁻¹⁰ N.

This force will act in upward direction as the force on the positive charge will always acts in the direction of the electric field.

Volume of rain droplet(V) = 4/3 π R³

Volume V = 4/3 x 3.14 x ( 1.25 x 10⁻³)³

                 = 6.54 × 10⁻⁹m³

As Density of water = 1000 kg / m³

Density =Mass/Volume

So, Mass of rain droplet = 1000 x 6.54 x 10⁻⁹ kg

= 6.54 x 10⁻⁶ kg .

Now, Weight(W) = mg

= 6.54 x 10⁻⁶ x 9.8

= 64.092 x 10⁻⁶ N.

As the electric force is 15 × 10⁻¹⁰ N

So, weight force is more than the electric force.

Hence, the magnitude of the electric force on the droplet is less as compare to the weight force.

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