Charges of +2.0, +3.0 and -8.0µC are placed at the vertices of an equilateral triangle of a side 10 cm. Calculate the magnitude of the force acting on the -8µC charge due to the other two loads.
F1 = 9x10^9 ( ((2x10^-6)(-8x10^-6))/ ( 0.1 )^2 ) = -14.4N
F2 = 9x10^9 ( ((3x10^-6)(-8x10^-6))/ ( 0.1 )^2 ) = -21.6N
F = √ (F1^2+F2^2+2(F1)(F2)(cos(θ)))
F = √ ((14.4)^2 + (21.6)^2 + 2(14.4)(21.6)(Cos(60)))
F = 31.38N
ArcTan((14.4Sen(60))/(21.6+14.4Cos(60)) = 23.41º
Case 2
F1 = 9x10^9 ( ((2x10^-6)(8x10^-6))/ ( 0.1 )^2 ) = 14.4N
F2 = 9x10^9 ( ((3x10^-6)(8x10^-6))/ ( 0.1 )^2 ) = 21.6N
Fex = 14.4Cos(120) - 21.6Cos(0) = -28.8
Fey = 14.4Sen(120) + 21.6Sen(0) = 12.47076581
Fe = √ ((28.8)^2+(12.47076581)^2) = 31.38407239N
ArcTan((14.4)/(21.6)) = 33.69006753
180 - 33.69006753 = 146.3099325º
Well, well, well. It seems like we have some electrifying charges and an equilateral triangle doing some electric dance moves!
To find the magnitude of the force acting on the -8µC charge, we'll have to whip out the good old Coulomb's Law. It states that the force between two charges is proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.
Now, let's break it down. We have two charges, +2.0µC and +3.0µC, and we're interested in the force on the -8.0µC charge. The important thing to remember is that the force is a vector, so it has both magnitude and direction.
First, we need to find the distance between the -8.0µC charge and the +2.0µC charge. Since they're at the vertices of an equilateral triangle, they form an angle of 60 degrees. The side length of the triangle is given as 10 cm. Using some trigon-trickery, we can find that the distance between the -8.0µC charge and the +2.0µC charge is 5 cm.
Now, let's calculate the force between the -8.0µC charge and the +2.0µC charge. Plugging the values into Coulomb's Law, we get:
F1 = k * |q1| * |q2| / r^2,
where k is Coulomb's constant (9 x 10^9 Nm²/C²), q1 is the magnitude of the -8.0µC charge, q2 is the magnitude of the +2.0µC charge, and r is the distance between them.
F1 = (9 x 10^9 Nm²/C²) * (8.0 x 10^-6 C) * (2.0 x 10^-6 C) / (0.05 m)^2
Now, repeat the above steps to find the force between the -8.0µC charge and the +3.0µC charge. We'll call it F2.
Finally, add the two forces together to find the net force on the -8.0µC charge. Remember to consider both magnitude and direction!
And there you have it! With a little bit of math and a whole lot of electric wit, you'll find the magnitude of the force acting on our -8.0µC charge. Keep those charges dancing, my friend!
To calculate the magnitude of the force acting on the -8µC charge, we can use Coulomb's law, which states that the force between two charges is given by:
F = k * |q1 * q2| / r^2
Where:
F is the force between the charges,
k is the Coulomb's constant (approximately 8.99 x 10^9 N m^2/C^2),
q1 and q2 are the magnitudes of the charges, and
r is the distance between the charges.
Let's calculate the force individually for each of the charges and then find the net force by taking the vector sum.
For the +2.0µC charge:
q1 = +2.0µC
q2 = -8.0µC
r = 10 cm = 0.1 m
Using Coulomb's law, the force between the +2.0µC and -8.0µC charges is:
F1 = k * |q1 * q2| / r^2
= 8.99 x 10^9 N m^2/C^2 * |(2.0 x 10^-6 C) * (-8.0 x 10^-6 C)| / (0.1 m)^2
Next, for the +3.0µC charge:
q1 = +3.0µC
q2 = -8.0µC
r = 10 cm = 0.1 m
Using Coulomb's law, the force between the +3.0µC and -8.0µC charges is:
F2 = k * |q1 * q2| / r^2
= 8.99 x 10^9 N m^2/C^2 * |(3.0 x 10^-6 C) * (-8.0 x 10^-6 C)| / (0.1 m)^2
Now, let's find the net force by taking the vector sum of the forces. Since the charges are arranged in an equilateral triangle, the forces will have equal magnitudes but will be in different directions.
The net force is given by:
F_net = √(F1^2 + F2^2 + 2 * F1 * F2 * cos(60°))
Substituting the values we calculated, we get:
F_net = √((F1^2 + F2^2 + 2 * F1 * F2 * cos(60°))
After performing the calculations, you will find the magnitude of the force acting on the -8µC charge due to the other two charges.
To calculate the magnitude of the force acting on the -8µC charge due to the other two charges, we can use Coulomb's Law:
F = k * (|q1| * |q2|) / r^2
Where:
F is the force between the charges,
k is the electrostatic constant (k = 8.99 x 10^9 N m^2/C^2),
|q1| and |q2| are the magnitudes of the charges, and
r is the distance between the charges.
In this case, we need to calculate the force acting on the -8µC charge.
Let's consider the charges as:
q1 = +2.0µC
q2 = +3.0µC
q3 = -8.0µC
Since the charges are placed at the vertices of an equilateral triangle, the distance between each charge can be calculated using trigonometry.
In an equilateral triangle, each side is equal to the length of the triangle's base. In this case, the side length is 10 cm.
Using the Law of Cosines, we can calculate the distance between the charges (r):
r = √(a^2 + b^2 - 2ab * cos(theta))
Where:
a = b = 10 cm (side length of the triangle)
theta = 60 degrees (angle between the sides of the triangle)
Now we can calculate the distance (r):
r = √(10^2 + 10^2 - 2 * 10 * 10 * cos(60))
= √(100 + 100 - 2 * 100 * cos(60))
= √(200 - 200 * 0.5)
= √(200 - 100)
= √100
= 10 cm
Now we have all the information needed to calculate the force acting on the -8µC charge:
F = k * (|q1| * |q3|) / r^2
F = (8.99 x 10^9 N m^2/C^2) * (2.0 x 10^-6 C) * (8.0 x 10^-6 C) / (10^-2 m)^2
F = (8.99 x 10^9) * (2.0 x 10^-6) * (8.0 x 10^-6) / (10^-4)
F = (8.99 x 2.0 x 8.0) * (10^-6 x 10^-6 / 10^-4)
F = (143.84) * (10^-6 / 10^-4)
F = 143.84 * 10^-6 x 10^4
F = 1438.4 N
Therefore, the magnitude of the force acting on the -8µC charge due to the other two charges is 1438.4 Newtons.