A small charge that has lost 5.0 X 10^3 electrons and has a mass of 2.0 x 10^-6 kg experiences an acceleration of 15ms^2 when placed at a point P in space. Determine the magnitude and direction of the electric field at P
15m/s^2
F = q E = m a
E = m a/q
m is given
q = 5*10^3 * electron charge
a is given
Why is 5*10^3 multiplied by a electron charge?
That is how much negative charge it is missing. (How much positive charge it has)
To determine the magnitude and direction of the electric field at point P, we can apply Newton's second law of motion.
1. Step 1: We need to determine the net force acting on the charge. The net force acting on a charged object is given by the equation F = q * E, where F is the net force, q is the charge, and E is the electric field.
2. Step 2: Rearrange the equation to solve for the electric field, E = F / q.
3. Step 3: Calculate the net force acting on the charge. The net force F can be determined by using Newton's law, F = m * a, where m is the mass of the charge and a is the acceleration.
4. Step 4: Plug in the given values and calculate the electric field magnitude.
Given:
Charge, q = -5.0 x 10^3 electrons = -5.0 x 10^3 * 1.6 x 10^-19 C (since 1 electron has a charge of 1.6 x 10^-19 C)
Mass, m = 2.0 x 10^-6 kg
Acceleration, a = 15 m/s^2
Calculations:
Step 1: F = m * a
F = (2.0 x 10^-6 kg) * (15 m/s^2)
F = 3.0 x 10^-5 N
Step 2: E = F / q
E = (3.0 x 10^-5 N) / (-5.0 x 10^3 * 1.6 x 10^-19 C)
E = -3.75 x 10^11 N/C
The magnitude of the electric field at point P is 3.75 x 10^11 N/C, and its direction is negative (-) since the charge lost electrons.
Note: In this explanation, I used the fundamental formulas and concepts related to electric fields, charges, and forces to solve the problem. Always make sure to check if the given values and units are compatible and pay attention to the direction of the charge to determine the sign of the electric field.