Magnitude and direction of charge

[flynostrich08]

Homework Statement

Two charges q1 = 5μC and q2 = 10μC are located at x = 0 cm and 10 cm respectively. What
would the magnitude and direction of the force on a 2μC charge located at x = 5 cm be due to charges q1 and q2.

Homework Equations

The Attempt at a Solution

i don't know how to do this problem, i appreciate any help, thanks

 

[pgardn]

Homework Statement

Two charges q1 = 5μC and q2 = 10μC are located at x = 0 cm and 10 cm respectively. What
would the magnitude and direction of the force on a 2μC charge located at x = 5 cm be due to charges q1 and q2.

Homework Equations

The Attempt at a Solution

i don't know how to do this problem, i appreciate any help, thanks

You don't have any equations you have learned about force and static charges?

 

[flynostrich08]

F=kq1q2/r^2 so do i find the force of each then subtract them? What do i do with each F i find?

i don't know any static charge equations.

 

[pgardn]

F=kq1q2/r^2 so do i find the force of each then subtract them? What do i do with each F i find?

i don't know any static charge equations.

What you have given above is the equation that says that two static charges q1 and q2 separated by a distance r, exert a force of F on each other. So why don't you start by determing the Force of the charge at the 0 cm mark on the charge that is in between, the 2uc charge.

 

[rwisz]

I would approach this problem by applying Coulomb's Law, which states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The equation for this is F = k*q1*q2/r^2, where k is the Coulomb's constant (9x10^9 Nm^2/C^2), q1 and q2 are the charges, and r is the distance between them.

In this problem, we have two charges (q1 = 5μC and q2 = 10μC) located at x = 0 cm and 10 cm respectively. The distance between them is 10 cm, or 0.1 m. We also have a third charge (2μC) located at x = 5 cm. To find the force on this charge due to q1 and q2, we can use the superposition principle, which states that the total force on a charge is the vector sum of the individual forces due to each charge.

First, let's find the force due to q1. Using the equation above, we have F1 = (9x10^9 Nm^2/C^2)*(5x10^-6 C)*(2x10^-6 C)/(0.05 m)^2 = 1.8x10^-2 N. This force will be directed towards q1, since the charges have the same sign (both positive).

Next, let's find the force due to q2. Using the same equation, we have F2 = (9x10^9 Nm^2/C^2)*(10x10^-6 C)*(2x10^-6 C)/(0.05 m)^2 = 3.6x10^-2 N. This force will be directed towards q2, since the charges have opposite signs (q2 is negative and our test charge is positive).

To find the net force on our test charge, we can add these two forces together as vectors. Since they are in opposite directions, the net force will be the difference between them. Therefore, the magnitude of the net force is F = |F2 - F1| = |3.6x10^-2 N - 1.8x10^-2 N| = 1.8x10^-2 N. The direction of this force will be towards q2, since