What is the magnitude of the electric field a distance 3.0mm from an electron?
Express your answer to two significant figures and include the appropriate units
same old formula
E = k Qe/r^2 N/C
Qe = - 1.6*10^-19 Coulombs
r = 3 * 10^-3 meters
k = 9 * 10^9 N m^2/C^2
E is toward negative charge
To find the magnitude of the electric field a distance 3.0 mm from an electron, you can use Coulomb's Law. Coulomb's Law states that the electric field produced by a point charge is proportional to the charge and inversely proportional to the square of the distance.
The formula for the electric field produced by a point charge is given by:
E = k * (q / r^2),
where E is the electric field, k is Coulomb's constant, q is the charge, and r is the distance.
In this case, the charge q is the charge of an electron, which is approximately -1.6 x 10^-19 C. The distance r is given as 3.0 mm, which can be converted to meters by multiplying it by 10^-3.
Now we can plug the values into the formula and calculate the electric field:
E = k * (-1.6 x 10^-19 C) / ((3.0 x 10^-3)^2)
The value of Coulomb's constant, k, is approximately 8.99 x 10^9 Nm^2/C^2.
Calculating the electric field:
E = (8.99 x 10^9 Nm^2/C^2) * (-1.6 x 10^-19 C) / ((3.0 x 10^-3)^2)
By plugging this into a calculator, you'll find that the magnitude of the electric field is approximately 3.2 x 10^12 N/C.
Rounded to two significant figures, the magnitude of the electric field is 3.2 x 10^12 N/C.