To find the magnitude of the electric field, we can use the formula:
Electric field strength = Force / Charge
In this case, the force acting on the water droplet is its weight, which can be calculated using the formula:
Weight = mass x acceleration due to gravity
The acceleration due to gravity is approximately 9.8 m/s^2 (assuming this is on the surface of the Earth). Therefore, the weight of the water droplet is:
Weight = 3.25 x 10^(-5) kg x 9.8 m/s^2 = 3.185 x 10^(-4) N
Now, we can calculate the magnitude of the electric field:
Electric field strength = Force / Charge
= 3.185 x 10^(-4) N / 4.00 x 10^(-15) C
≈ 7.96 x 10^(10) N/C
Therefore, the magnitude of the electric field is approximately 7.96 x 10^(10) N/C.
On a stormy day, a water droplet has a mass of 3. 25 x 10 kg is suspended in the air above ground due to an atmospheric electric field. -15 The water droplet has a charge of 4,00 x 10 C. The atmospheric electric field points vertically downward and has an unknown magnitude. Find the magnitude of this electric field
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