A small object of mass 3.50 g and charge -22 µC "floats" in a uniform electric field. What is the magnitude and direction of the electric field?
E q + m g = 0
To find the magnitude and direction of the electric field, you can use the equation:
F = qE
where F is the electrostatic force experienced by the object, q is the charge of the object, and E is the electric field strength.
In this case, the object is in equilibrium or "floating," which means the electrostatic force acting on it is balanced by other forces such as gravitational force.
Since the object is floating, we know that the electrostatic force (Fe) is equal in magnitude but opposite in direction to the gravitational force (Fg).
Taking the magnitude of the charge (q) to be the absolute value of -22 µC, or 22 x 10^-6 C, we can rewrite the equation as:
|Fe| = |Fg|
Using Newton's second law (F = mg) to find the gravitational force, where m is the mass of the object, and g is the acceleration due to gravity (approximately 9.8 m/s^2), we have:
|Fe| = |m * g|
Substituting the given mass (3.50 g) into the equation, we find:
|Fe| = (3.50 g) * (9.8 m/s^2)
|Fe| = 34.3 x 10^-3 N
Now, since the electrostatic force (Fe) is equal in magnitude and opposite in direction to the gravitational force (Fg), the magnitude of the electric field (E) can be found by rearranging the equation:
|E| = |Fe / q|
Substituting the values we found, we get:
|E| = (34.3 x 10^-3 N) / (22 x 10^-6 C)
|E| = 1.56 x 10^3 N/C
The magnitude of the electric field is 1.56 x 10^3 N/C.
To determine the direction of the electric field, we need additional information. If the charge on the object is negative (-22 µC), it will experience a force in the opposite direction to the electric field. Therefore, in this case, the direction of the electric field would be opposite to the direction in which the object is "floating."