a +35x10^-6 C point charge is placed 32 cm from an identical +32x10^-6 C charge. how much work would be required to move a +50.0x10^-6 C test charge from a point midway between them to a point 12 cm closer to either of the charges?
please show work and explain. i still don't understand what Damon said. we didn't learn how to solve it with integrals. is there another way
physics - Damon, Friday, February 22, 2008 at 4:03pm
q is each of our two charges
Q is our test charge
Left charge at x = 0
Right charge at x = .32 m
force due to left charge = k q Q/x^2
force due to right charge = -k q Q/(.32-x)^2
when x = .16, the middle, the sum of those two forces is zero. However as you move off center, the one nearer will push back harder. We move to x = .16+.12 = .28. At that point we are .32 -.28 = .04 from the right charge.
The integral of dr/r^2 = -1/r =-[1/Rend -1/Rbegin] = [1/Rbegin - 1/Rend]
Let's assume we move right (+x direction)
Work done against left charge (negative because it is pushing in the direction of motion so we are holding back) = - k q Q( 1/.16 -1/.28)
Work done against right charge = +k q Q(1/.04 -1/.16)
If you did not have the integrals to derive the equation, I bet you were given an equation like this:
Potential Energy of charge close to other charge = PE or U = k Q1 Q2 / R
That is all I did with the integrals.
That is where the 1/Rend and 1/Rbegin comes from.
The work done is just the change in potential energy.
Since I knew how to integrate, I did the force times distance, but I bet the result of my integral, the 1/R equation, is in your book.
1.3
To find the work required to move the test charge from a point midway between the charges to a point 12 cm closer to either of the charges, we can use the concept of work done against electric forces.
Let's break down the steps to find the work done against each charge individually:
1. Work done against the left charge (q1):
The force exerted by the left charge on the test charge at position x is given by:
Force1 = k * q * Q / x^2
We need to find the work done to move the test charge from x = 0.16 m to x = 0.28 m against this force. We can use the formula for work by integrating the force over the displacement:
Work1 = ∫(-Force1) * dx
Work1 = - ∫(k * q * Q / x^2) * dx (integrating from 0.16 to 0.28)
2. Work done against the right charge (q2):
The force exerted by the right charge on the test charge at position x is given by:
Force2 = -k * q * Q / (0.32 - x)^2
We need to find the work done to move the test charge from x = 0.16 m to x = 0.28 m against this force. Again, we can use the formula for work:
Work2 = ∫(-Force2) * dx
Work2 = ∫(k * q * Q / (0.32 - x)^2) * dx (integrating from 0.12 to 0.16)
Now, let's calculate the work done against each charge:
Work1 = - k * q * Q * (∫(1/x^2) * dx) (integrating from 0.16 to 0.28)
Work1 = - k * q * Q * (-1/x) |_0.16^0.28
Work1 = k * q * Q * (1/0.28 - 1/0.16)
Work2 = k * q * Q * (∫(1/(0.32 - x)^2) * dx) (integrating from 0.12 to 0.16)
Work2 = k * q * Q * (1/(0.32 - x)) |_0.12^0.16
Work2 = k * q * Q * (1/0.04 - 1/0.16)
Finally, we can plug in the given values of the charges:
q = 35x10^-6 C, Q = 50.0x10^-6 C
Work1 = (9 * 10^9 N m^2/C^2) * (35x10^-6 C) * (50.0x10^-6 C) * (1/0.28 - 1/0.16)
Work2 = (9 * 10^9 N m^2/C^2) * (35x10^-6 C) * (50.0x10^-6 C) * (1/0.04 - 1/0.16)
Calculating these values will give you the work done against each charge. Remember that the work done against the left charge has a negative sign because we are holding back against its force of repulsion. The work done against the right charge has a positive sign since we are moving in the direction of its force of attraction.
I hope this explanation helps you understand the process of calculating the work required in this scenario.