Three charges sit on the vertices of an equilateral triangle, the side of which are 30.0 cm long. If the charges are A = +4.0 C, B = +5.0 C, and C = +6.0 C (clockwise from the top vertex), find the force on each charge.
physics - sara, Friday, August 17, 2012 at 10:45pm
How did you get F12 and F13 ??
F12 = k•q1•q2/a²=9•10^9•4•10^-6•5•10^-6/0.09=2 N
F13 = k•q1•q3/a² = 9•10^9•4•10^-6•6•10^-6/0.09=2.4 N
F(A) = sqrt(F12² + F13² - 2•F12•F13•cos 120º)=
=sqrt[4+5.76-2•2•2.4(-0.5)] =3.82 N
Charge A(q1)
F12 = k•q1•q2/a².
F13 = k•q1•q3/a².
F(A) = sqrt(F12² + F13² - 2•F12•F13•cos 120º).
Charge B(q2)
F21 = k•q1•q2/a².
F23 = k•q2•q3/a².
F(B) = sqrt(F21²+F23²-2•F21•F23•cos120º).
Charge C (g3),
F31 = k•q1•q3/a².
F32 = k•q2•q3/a².
F(C) = sqrt(F31² + F32² - 2•F31•F32•cos 120º).
How did you get F12 and F13 ??
same question above, but calculate the magnitude of the net force that each charge experiences.
To find the force on each charge, we will use Coulomb's Law, which states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. The equation for Coulomb's Law is:
F = (k * |q1 * q2|) / r^2
Where:
F is the force between the charges.
k is the electrostatic constant, which is approximately 8.99 x 10^9 N·m^2/C^2.
q1 and q2 are the magnitudes of the charges involved.
r is the distance between the charges.
Let's calculate the forces on each charge:
1. Force on charge A (4.0 μC):
- Charge A attracts charge B and charge C.
- The distance between A and B is equal to the length of the side of the equilateral triangle: 30.0 cm.
- The distance between A and C is also equal to the length of the side of the equilateral triangle: 30.0 cm.
- Using Coulomb's Law, the force between A and B is given by:
FAB = (k * |qA * qB|) / r^2
FAB = (8.99 x 10^9 N·m^2/C^2) * (4.0 μC * 5.0 μC) / (0.30 m)^2
FAB = 11.98 N
- Using Coulomb's Law, the force between A and C is given by:
FAC = (k * |qA * qC|) / r^2
FAC = (8.99 x 10^9 N·m^2/C^2) * (4.0 μC * 6.0 μC) / (0.30 m)^2
FAC = 14.38 N
2. Force on charge B (5.0 μC):
- Charge B attracts charge C.
- The distance between B and C is equal to the length of the side of the equilateral triangle: 30.0 cm.
- Using Coulomb's Law, the force between B and C is given by:
FBC = (k * |qB * qC|) / r^2
FBC = (8.99 x 10^9 N·m^2/C^2) * (5.0 μC * 6.0 μC) / (0.30 m)^2
FBC = 16.77 N
3. Force on charge C (6.0 μC):
- Charge C is attracted by both charge A and charge B.
- The distances between C and A and between C and B are equal to the length of the side of the equilateral triangle: 30.0 cm.
- Using Coulomb's Law, the force between C and A is given by:
FCA = (k * |qC * qA|) / r^2
FCA = (8.99 x 10^9 N·m^2/C^2) * (6.0 μC * 4.0 μC) / (0.30 m)^2
FCA = 17.97 N
- Using Coulomb's Law, the force between C and B is given by:
FCB = (k * |qC * qB|) / r^2
FCB = (8.99 x 10^9 N·m^2/C^2) * (6.0 μC * 5.0 μC) / (0.30 m)^2
FCB = 22.47 N
Therefore, the forces on each charge are:
- Force on charge A (4.0 μC): FAB = 11.98 N, FAC = 14.38 N
- Force on charge B (5.0 μC): FBC = 16.77 N
- Force on charge C (6.0 μC): FCA = 17.97 N, FCB = 22.47 N