Anonymous
Jul 25
The charges are located on the x- axis of a number plane as follows, Q1=-0.15 micron C at x = 0, q2=+1.5micronC at x=3.0 m and q3=+2.12micronC at x=4.5m.Calculate the magnitude and direction of the ?
Electrostatic force on charge q2?
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ChatGPT
Jul 25
net force on a charge of +3.0 micron C located at x = 1.5 m.
To calculate the net force on the charge at x = 1.5 m, we need to consider the forces due to each of the charges Q1, q2, and q3.
The force between two charges can be calculated using Coulomb's law:
F = k * |q1| * |q2| / r^2
where k is the Coulomb constant (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.
First, let's calculate the force due to Q1 on the charge at x = 1.5 m:
r1 = 1.5 m
F1 = k * |Q1| * |q| / r1^2
F1 = (8.99 x 10^9) * (0.15 x 10^-6) * (3.0 x 10^-6) / (1.5)^2
F1 = 0.18 N
Next, let's calculate the force due to q2 on the charge at x = 1.5 m:
r2 = 1.5 m
F2 = k * |q2| * |q| / r2^2
F2 = (8.99 x 10^9) * (1.5 x 10^-6) * (3.0 x 10^-6) / (1.5)^2
F2 = 0.36 N
Finally, let's calculate the force due to q3 on the charge at x = 1.5 m:
r3 = 3.0 m
F3 = k * |q3| * |q| / r3^2
F3 = (8.99 x 10^9) * (2.12 x 10^-6) * (3.0 x 10^-6) / (3.0)^2
F3 = 0.71 N
Now, we can calculate the net force on the charge at x = 1.5 m by summing the forces due to each charge:
Fnet = F1 + F2 + F3
Fnet = 0.18 N + 0.36 N + 0.71 N
Fnet = 1.25 N
Therefore, the magnitude of the net force on the charge at x = 1.5 m is 1.25 N. To find the direction of the force, we need to consider the direction of each individual force and sum them up vectorially.
To calculate the net force on the charge at x = 1.5 m, we need to consider the forces due to each of the charges Q1, q2, and q3.
The force between two charges can be calculated using Coulomb's law:
F = k * |q1| * |q2| / r^2
where k is the Coulomb constant (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.
First, let's calculate the force due to Q1 on the charge at x = 1.5 m:
r1 = 1.5 m
F1 = k * |Q1| * |q| / r1^2
F1 = (8.99 x 10^9) * (0.15 x 10^-6) * (3.0 x 10^-6) / (1.5)^2
F1 = 0.18 N
Next, let's calculate the force due to q2 on the charge at x = 1.5 m:
r2 = 1.5 m
F2 = k * |q2| * |q| / r2^2
F2 = (8.99 x 10^9) * (1.5 x 10^-6) * (3.0 x 10^-6) / (1.5)^2
F2 = 0.36 N
Finally, let's calculate the force due to q3 on the charge at x = 1.5 m:
r3 = 3.0 m
F3 = k * |q3| * |q| / r3^2
F3 = (8.99 x 10^9) * (2.12 x 10^-6) * (3.0 x 10^-6) / (3.0)^2
F3 = 0.71 N
Now, we can calculate the net force on the charge at x = 1.5 m by summing the forces due to each charge:
Fnet = F1 + F2 + F3
Fnet = 0.18 N + 0.36 N + 0.71 N
Fnet = 1.25 N
Therefore, the magnitude of the net force on the charge at x = 1.5 m is 1.25 N. To find the direction of the force, we need to consider the direction of each individual force and sum them up vectorially.
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Google Gemini AI
Jul 25
The electric field due to a point charge q at a distance r from it is given by
$$E= k\frac{q}{r^2}$$
where k is the coulombs constant.
The electric field due to q1 at q2 is
$$E_1 = k\frac{q_1}{(3\text{ m})^2}$$
The electric field due to q3 at q2 is
$$E_3 = k\frac{q_3}{(4.5\text{ m})^2}$$
The total electric field at q2 is
$$E_{net}=E_1+E_3=k\frac{q_1}{(3\text{ m})^2}+k\frac{q_3}{(4.5\text{ m})^2}$$
$$E=9\times10^9\frac{(-0.15\times10^{-6})\text{ C}}{(3\text{ m})^2}+9\times10^9\frac{(2.12\times10^{-6})\text{ C}}{(4.5\text{ m})^2}$$
$$E=-3.375\times10^4\frac{\text{ N}}{\text{ C}}$$
The negative sign indicates that the electric field is in the -x direction.
Therefore, the magnitude of the electric field at q2 is
$$|E|=3.375\times10^4\frac{\text{ N}}{\text{ C}}$$
and the direction is in the -x direction.
$$E= k\frac{q}{r^2}$$
where k is the coulombs constant.
The electric field due to q1 at q2 is
$$E_1 = k\frac{q_1}{(3\text{ m})^2}$$
The electric field due to q3 at q2 is
$$E_3 = k\frac{q_3}{(4.5\text{ m})^2}$$
The total electric field at q2 is
$$E_{net}=E_1+E_3=k\frac{q_1}{(3\text{ m})^2}+k\frac{q_3}{(4.5\text{ m})^2}$$
$$E=9\times10^9\frac{(-0.15\times10^{-6})\text{ C}}{(3\text{ m})^2}+9\times10^9\frac{(2.12\times10^{-6})\text{ C}}{(4.5\text{ m})^2}$$
$$E=-3.375\times10^4\frac{\text{ N}}{\text{ C}}$$
The negative sign indicates that the electric field is in the -x direction.
Therefore, the magnitude of the electric field at q2 is
$$|E|=3.375\times10^4\frac{\text{ N}}{\text{ C}}$$
and the direction is in the -x direction.
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Abenezer Tesfaye
Jul 25
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Abenezer Tesfaye
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