A 10 charge sits at a point in space where the magnitude of the electric field is 1900 .What will the magnitude of the field be if the 10 charge is replaced by a 20 charge?
1800 N/C
To find the magnitude of the electric field when the 10 C charge is replaced by a 20 C charge, we can start by using Coulomb's law.
Coulomb's law states that the magnitude of the electric field, E, at a certain point in space is equal to the electrostatic force, F, experienced by a test charge, q, placed at that point, divided by the magnitude of the test charge:
E = F / q
In this case, the test charge is positive and has a magnitude of 1 C, so we can simplify the equation to:
E = F
Since the electric field is a vector quantity, we're only interested in its magnitude, so we don't need to consider the direction.
The electrostatic force between two charges can be calculated using the formula:
F = (k * |q1| * |q2|) / r^2
Where:
- k is the electrostatic constant, approximately equal to 9 x 10^9 Nm^2/C^2
- |q1| and |q2| are the magnitudes of the charges
- r is the distance between the charges
In this case, we have a 10 C charge at a point in space, and we want to find the electric field when it is replaced by a 20 C charge. Since we're only interested in the magnitude of the electric field, we can plug in the magnitudes of the charges:
F = (k * 10 C * 20 C) / r^2
Now, if we compare this equation to the one we started with (E = F), we can see that the magnitude of the electric field is directly proportional to the force. Therefore, when we double the charge (from 10 C to 20 C), the magnitude of the electric field will also double, as the force doubles.
Therefore, the magnitude of the electric field will be 2 * 1900 N/C, which is equal to 3800 N/C when the 10 C charge is replaced by a 20 C charge.