Question
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
Answers
GPT 3.5
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.
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.