What is the electric field at a distance of 2 cm from 11 mC of negative charge?
2cm converts to 0.02 meters
E = k•|q|/r²,
k =9•10^9 N•m^2/C^2,
|q| =11•10^-3 C,
r =0.02 m
Field lines are directed to the charge
To calculate the electric field at a given distance from a charge, you can use Coulomb's Law. Coulomb's Law states that the electric field created by a point charge is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance from the charge.
The formula for the electric field (E) created by a point charge is given by:
E = k * (Q / r^2)
Where:
E is the electric field,
k is the electrostatic constant (approximately 9 x 10^9 Nm^2/C^2),
Q is the magnitude of the charge, and
r is the distance from the charge.
In this case, the magnitude of the charge is 11 mC, which can be converted to 11 x 10^-3 C.
Plugging in the values into the formula, we have:
E = (9 x 10^9 Nm^2/C^2) * (11 x 10^-3 C) / (0.02 m)^2
Simplifying,
E = (9 x 10^9 Nm^2/C^2) * (11 x 10^-3 C) / (0.0004 m^2)
E = 99 x 10^6 N/C
Therefore, the electric field at a distance of 2 cm (0.02 meters) from an 11 mC negative charge is 99 x 10^6 N/C.