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
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.
To find the magnitude of the electric field, we can use the equation:
Electric field (E) = Force (F) / Charge (Q)
The force experienced by the water droplet can be calculated using the equation:
Force (F) = Mass (m) * Acceleration due to gravity (g)
Given:
Mass (m) = 3.25 x 10^-5 kg
Acceleration due to gravity (g) = 9.8 m/s^2
Charge (Q) = 4.00 x 10^-5 C
First, let's calculate the force experienced by the water droplet:
F = m * g
F = 3.25 x 10^-5 kg * 9.8 m/s^2
F = 3.18 x 10^-4 N
Now, we can substitute the obtained force and the given charge into the formula to find the electric field:
E = F / Q
E = 3.18 x 10^-4 N / 4.00 x 10^-5 C
E = 7.95 N/C
Therefore, the magnitude of the electric field is 7.95 N/C.
To find the magnitude of the atmospheric electric field, we can use the equation:
F = qE
where F is the force experienced by the charged droplet, q is the charge of the droplet, and E is the magnitude of the electric field.
In this problem, the force experienced by the droplet is due to the gravitational force acting downwards and the electric force acting upwards. So we can write the equation as:
mg - Eq = 0
where m is the mass of the droplet, g is the acceleration due to gravity, and E is the magnitude of the electric field.
Rearranging the equation, we get:
Eq = mg
Now we can substitute the given values into the equation. The mass of the droplet is 3.25 x 10^(-5) kg and the acceleration due to gravity is approximately 9.8 m/s^2. The charge of the droplet is 4.00 x 10^(-19) C.
So the equation becomes:
(4.00 x 10^(-19) C)E = (3.25 x 10^(-5) kg)(9.8 m/s^2)
Next, we can solve for the magnitude of the electric field E by dividing both sides of the equation by the charge of the droplet:
E = [(3.25 x 10^(-5) kg)(9.8 m/s^2)] / (4.00 x 10^(-19) C)
Evaluating this expression gives us the magnitude of the atmospheric electric field.