A 10C charge is positioned at x=0m, a -5C charge is positioned at x=0.06m and a 2C charge is at 0.09m. What is the net force on the -5C charge? answer is (Fnet = -2.5x1013 N)
10c-<0.06>=f(n)(l1/l3+l2/l4)
uf=f(x)_ln<1.962>
fx/%fl
%fl=1.333
2C-<0.09>=f(1.33)
fnet=-2.5x1013N
To calculate the net force on the -5C charge, we need to consider the electric forces exerted by the other charges.
The electric force between two charges can be calculated using Coulomb's Law:
F = k * (|q1| * |q2|) / r^2
Where:
F is the electric force
k is the electrostatic constant (k = 9 * 10^9 Nm^2/C^2)
|q1| and |q2| are the magnitudes of the charges
r is the distance between the charges
Let's calculate the electric force between the -5C charge and the 10C charge. The distance between them is 0.06m:
F1 = (9 * 10^9 Nm^2/C^2) * (|-5C| * |10C|) / (0.06m)^2
= (9 * 10^9 Nm^2/C^2) * (5C * 10C) / 0.0036m^2
= 450 * 10^6 N / 0.0036m^2
= 1.25 * 10^14 N
The magnitudes of the charges are used in this calculation because the force is always positive, and the direction is determined by the signs of the charges.
Now let's calculate the electric force between the -5C charge and the 2C charge. The distance between them is 0.03m:
F2 = (9 * 10^9 Nm^2/C^2) * (|-5C| * |2C|) / (0.03m)^2
= (9 * 10^9 Nm^2/C^2) * (5C * 2C) / 0.0009m^2
= 90 * 10^6 N / 0.0009m^2
= 1 * 10^14 N
Since the 10C and 2C charges have positive values, the net force on the -5C charge is in the opposite direction of the resultant forces calculated above. The net force Fnet on the -5C charge is given by:
Fnet = F1 - F2
= (1.25 * 10^14 N) - (1 * 10^14 N)
= 0.25 * 10^14 N
= -2.5 * 10^13 N
Therefore, the net force on the -5C charge is -2.5 * 10^13 N.