Due to electric effects on the earth and in the atmosphere, there is an electric field near the earth's surface that points down. Assuming the electric field is uniform with a magnitude of 150 N/C, what charge would a 145g baseball need to float as though it were weightless?

I am quite confused by what formula I need to apply to this, and what condition it needs to meet. The ball needs to acquire enough charge to overcome a 150 N/C electric field....but how to represent that numerically as one side of an equation?

force gravity=electric force

9.8N/kg*massball=Eq
q= 9.8*massball/E=
massball=.145kg
so q= ...

Electric charge is distributed uniformly along an infinitely long, thin wire as shown in figure below. The linear charge density λ is 1.5 cross times 10 to the power of negative 10 end exponent space C divided by m.

q14.png

Find the magnitude of electric field at 108 cm from the infinitely long wire.

Well, if the baseball needs to acquire enough charge to overcome a 150 N/C electric field, we can start by considering the electric force on the baseball.

The electric force (F) on an object is given by the equation F=qE, where q is the charge of the object and E is the electric field.

Since we want the baseball to be weightless, the electric force (F) needs to cancel out the gravitational force (mg) acting on it, where m is the mass of the baseball and g is the acceleration due to gravity.

So we can set up the equation mg = qE.

Now, we know that the mass of the baseball is 145g which is equivalent to 0.145 kg, and the electric field is given as 150 N/C.

Plugging these values into the equation, we have:

0.145 kg * g = q * 150 N/C.

Now, we need to solve for the charge (q). But hey, let's not get too charged up about it and light a bulb on our head. Instead, why not solve this equation together?

First, we can divide both sides of the equation by 150 N/C:

(0.145 kg * g) / 150 N/C = q.

Now, we just need to know the value of g (the acceleration due to gravity). Do you happen to remember this value?

To determine the charge required for the baseball to float as though it were weightless, you can use the formula:

F = q * E

Where:
- F is the force experienced by the charged baseball (opposite to the direction of the electric field)
- q is the charge of the baseball
- E is the magnitude of the electric field

In this case, the force experienced by the baseball needs to be equal in magnitude and opposite in direction to its weight (mg), which is given by Newton's second law:

F = mg

Substituting this value for F in the equation above, we have:

mg = q * E

Now, let's solve for q:

q = mg / E

Given:
m = 145 g = 0.145 kg
E = 150 N/C

Substituting these values into the equation, we get:

q = (0.145 kg * 9.8 m/s^2) / 150 N/C

Calculating this expression, we find:

q ≈ 9.42667 × 10^-3 C

So, the baseball would need to acquire a charge of approximately 0.00943 C in order to float as though it were weightless in a uniform electric field of magnitude 150 N/C.

To solve this problem, we can use the concept of electric force. The electric force on an object can be calculated using the formula:

F = qE

Where:
F is the magnitude of electric force
q is the charge of the object
E is the magnitude of the electric field

In this case, we want to find the charge (q) that would make the baseball float, meaning the electric force would balance its gravitational force.

The weight (W) of the baseball can be calculated using the formula:

W = mg

Where:
m is the mass of the baseball (145g in this case)
g is the acceleration due to gravity (approximately 9.8 m/s^2)

Since the baseball is weightless when it floats, the electric force must balance the weight, so we have:

F = W

Now we can substitute the formulas for electric force and weight:

qE = mg

Rearranging the equation, we can solve for q:

q = (mg) / E

First, we need to convert the mass of the baseball from grams to kilograms:

m = 145g = 0.145kg

Now we can plug in the given values:

q = (0.145kg * 9.8 m/s^2) / 150 N/C

Calculating this equation will give you the value for the charge (q) that the baseball needs to acquire in order to float as though it were weightless.