Charged Spheres Placed in an Electric Field, Suspended by Strings

[TheSoftAttack]

Homework Statement

Two small spheres, each of mass 4.00g, are suspended by light strings 15cm in length. A uniform electric field is applied in the x direction. The spheres have charges equal to -7*10^-8 C and +7*10^-8 C. Determine the electric field that enables the spheres to be in equilibrium at an angle of $$\theta$$ = 13$$^{o}$$.

Homework Equations

F = $$\frac{k_{e}Q_{1}Q_{2}}{r^{2}}$$
F = mgsin$$\theta$$

The Attempt at a Solution

So, for this problem I thought it would be a good idea to calculate the force of the two charged spheres attracting each other. This is given by the equation F = $$\frac{k_{e}Q_{1}Q_{2}}{r^{2}}$$. We already know what Q1 and Q2 are, and k is a constant. To calculate r, we just take the area between the two strings, and divide it into two right triangles. That way, we can use the Pythagorean theorem to get that (r/2) = (The length of the string)(sin$$\theta$$) = 0.15sin13. Now we have:

F = (8.99*10^9)(-7*10^-8)(7*10^-8)/((2*0.15sin13)^2) = -0.00967N

I also calculated the horizontal component of the gravitational force, which is given by F=mgsin$$\theta$$. This gives us (4*10^-3)(9.8)(sin13) = 0.00882N

From the perspective of the negatively charged sphere, both of these forces are acting in the same direction (pulling the sphere to the right), so I added the absolute value of these two quantities to get (0.00967 + 0.00882) = 0.01849N

Because the horizontal forces on the charged sphere need to add up to zero for the sphere to be in equilibrium, there needs to also be a force of 0.01849N pushing the sphere to the left. This force will be caused by the uniform electric field.

To get the magnitude of this electric field, we use the equation F = qE.

E = F/q = (-0.01849N)/(-7*10^-8) = 264,151 N/C.

This is the answer I got, but it's not correct. If anyone knows what I am doing incorrectly / what I actually need to be doing, I'd appreciate any help.

 

[Doc Al]

I also calculated the horizontal component of the gravitational force, which is given by F=mgsin$$\theta$$. This gives us (4*10^-3)(9.8)(sin13) = 0.00882N

Gravity acts vertically--it has no horizontal component.

Because the horizontal forces on the charged sphere need to add up to zero for the sphere to be in equilibrium

True. Note that vertical forces must also add to zero.

Hint: Don't forget about the tension in the string.

 

[TheSoftAttack]

Gravity acts vertically--it has no horizontal component.

Note that vertical forces must also add to zero.

Hint: Don't forget about the tension in the string.

Ok, so I thought about it more, and came up with something which is hopefully a little closer to the actual answer (it still isn't right though :( )

I'll just reiterate the variables we have:
$$\theta$$ = 13$$^{o}$$
m = 4*10^-3 kg
g = 9.8 m/s^2
Q1 = -7*10^-8 C
Q2 = 7*10^8 C
r = 0.3sin13$$^{o}$$
k = 8.99*10^9 Nm^2/s^2
T unknown
E unknown
Solve for E.

Because both the horizontal and vertical forces must add to zero, we can use the following two equations, taken from the perspective of the negatively charged sphere to the left:

$$\sum$$Fx = Tsin$$\theta$$ + k(Q1)(Q2)/r^2 - Q1E = 0

$$\sum$$Fy = Tcos$$\theta$$ -mg = 0

Solving the second equation for T, we get T = mg/cos$$\theta$$, and plugging back into the first equation, we get:

mgtan$$\theta$$+ k(Q1)(Q2)/r^2 = Q1E

Now, adding in the numbers:

(4*10^-3)(9.8)(tan13) + (8.99*10^9)(-7*10^-8)(7*10^-8)/(.3sin13)^2 = (-7*10^-8)E

0.00905 -0.00290 = (-7*10^-8)E

E = (0.00615)/(-7*10^-8) = -87.9 kN/C

Any clue as to what is wrong now? :/

 

[Doc Al]

Solving the second equation for T, we get T = mg/cos$$\theta$$, and plugging back into the first equation, we get:

mgtan$$\theta$$+ k(Q1)(Q2)/r^2 = Q1E

Looks OK.

Now, adding in the numbers:

(4*10^-3)(9.8)(tan13) + (8.99*10^9)(-7*10^-8)(7*10^-8)/(.3sin13)^2 = (-7*10^-8)E

Careful! Leave out the signs of the charges when you're finding the magnitudes of the forces.

Use the signs to tell you the direction of the forces and fields, but don't just plug the signs into the formulas and expect that to work. For example. The force between the charges is attractive, thus points to the right and is a positive force. But if you plug the signs in blindly, you'll get (+)(-) = (-), thus a negative force. Which gives you the wrong answer, of course.

(You could use the formulas blindly if they were written in vector form, but that's more of a pain than it's worth.)