Four charges are placed at the four corners of a square of side 15 cm. The charges on the upper left and right corners are +3 µC and -6 µC respectively. The charges on the lower left and right corners are -2.4 µC and -9 µC respectively. The net electric force on -6 µC charge is:
a) 25 N, 780 above the -ve x- axis
b) 25 N, 780 below the -ve x- axis
c) 13 N, 310 below the +ve x- axis
d) 18 N, 750 above the +ve x- axis
Thanks!!!!!!
F = k q1 q2 /r^2
all on 6
F = (k q6) q/r^2
let d = .15 meter
due to 3
Fx = - (k q6) 3/d^2 negative direction because +and -
Fy = 0
due to -2.4 repelling
Fx = +(kq6) 2.4 /[ 2*(d)^2](sqrt2/2)
Fy = same up
due to -1 only up
Fy = (kq6)9/ d^2
so
Fx = -c 3/d^2 + c 2.4 *sqrt 2/4d^2
Fy = +c 9 /d^2 + c 2.4 *sqrt 2/4d^2
Fx = (c/d^2) [(2.4/4) sqrt 2 -3]
Fy = (c/d^2) [(2.4/4) sqrt 2+9 ]
tan angle = Fy/Fx
= [.848 + 9]/[.848 -3]
= [ 9.848 /-2.15]
= - 4.58
overall being pushed up and pulled left so
so angle = 77.7 degree above -x axis
I will leave you to do the magnitude
but it is about 78 degrees above the -x axis :)
so if by 780 you mean 78, the answer is a
To find the net electric force on the -6 µC charge, we need to calculate the electric force exerted by each of the other charges and then add them up.
Let's label the charges:
Q1 = +3 µC (upper left corner)
Q2 = -6 µC (upper right corner)
Q3 = -2.4 µC (lower left corner)
Q4 = -9 µC (lower right corner)
First, let's calculate the electric force between the -6 µC and +3 µC charges.
We can use Coulomb's Law to do this:
F1 = k * |Q1 * Q2| / r^2
where k is the electrostatic constant (9 * 10^9 N m^2/C^2), |Q1 * Q2| is the absolute value of the product of the charges, and r is the distance between the charges.
The distance between -6 µC and +3 µC charges is the side length of the square, which is 15 cm or 0.15 m.
Plugging in the values:
F1 = (9 * 10^9 N m^2/C^2) * |(-6 µC) * (3 µC)| / (0.15 m)^2
Next, let's calculate the electric force between the -6 µC and -2.4 µC charges.
Using Coulomb's Law:
F2 = k * |Q2 * Q3| / r^2
The distance between -6 µC and -2.4 µC charges is also 0.15 m.
Plugging in the values:
F2 = (9 * 10^9 N m^2/C^2) * |(-6 µC) * (-2.4 µC)| / (0.15 m)^2
Finally, let's calculate the electric force between the -6 µC and -9 µC charges:
F3 = k * |Q2 * Q4| / r^2
Again, the distance between -6 µC and -9 µC charges is 0.15 m.
Plugging in the values:
F3 = (9 * 10^9 N m^2/C^2) * |(-6 µC) * (-9 µC)| / (0.15 m)^2
Now, we can find the net electric force on the -6 µC charge by adding the forces F1, F2, and F3 together.
Net electric force = F1 + F2 + F3
After calculating this value, we can compare it to the answer choices provided to determine the correct option.