A small object of mass 2.60 g and charge -17 µC "floats" in a uniform electric field. What is the magnitude and direction of the electric field?
Solve
Q*E = M*g
to get the field, E. The units will be Newtons per Coulomb
The direction of the E field must be down, so that the electrostatic force on the negative particle will be up.
2.60*17
To find the magnitude and direction of the electric field, we can use the equation F = qE, where F is the electrostatic force acting on the object, q is the charge of the object, and E is the electric field.
Since the object "floats," it means that the electrostatic force acting on it is exactly balanced by another force (such as gravity or tension in a string). In this case, we can assume that the gravitational force is negligible compared to the electrostatic force.
So, let's start by calculating the electrostatic force acting on the object using the given mass and charge.
Step 1: Convert the mass from grams to kilograms:
Mass = 2.60 g = 2.60 × 10^(-3) kg
Step 2: Substitute the given values into the equation for the electrostatic force:
F = qE
F = (-17 × 10^(-6) C) × E
Step 3: Set the electrostatic force equal to zero since the object "floats":
0 = (-17 × 10^(-6) C) × E
Now, solve for the electric field (E):
E = 0 / (-17 × 10^(-6) C)
E = 0 N/C
The magnitude of the electric field is 0 N/C, which means the electric field is effectively zero.
Since the electrostatic force is balanced and the object is not experiencing any net force, we cannot determine the direction of the electric field in this case.