# Need help solving electric forces and fields

1. Sep 5, 2009

### Anthem26

These are the problems I have left to do, but I have no clue as to how to solve them.

1. The problem statement, all variables and given/known data
a. Two small metal spheres are 23.0 cm apart. The spheres have equal amounts of negative charge and repel each other with a force of 0.040 N. What is the charge on each sphere?

b.A tiny sphere with a charge of q = +8.2 µC is attached to a spring. Two other tiny charged spheres, each with a charge of -4.0 µC, are placed in the positions shown in the figure, in which b = 4.2 cm. The spring stretches 5.0 cm from its previous equilibrium position toward the two spheres. Calculate the spring constant. (diagram: http://imgur.com/FKcPp.gif )

c.What are the magnitude and direction of the electric field midway between two point charges, -13 µC and +15 µC, that are 7.3 cm apart?

d.Two equal charges (Q = +0.95 nC) are situated at the diagonal corners A and B of a square of side x = 1.0 m as shown in the diagram. What is the magnitude of the electric field at point D? (diagram: http://imgur.com/eervv.gif )

2. Relevant equations
For all problems, I used Coulumb's law: F = k|q1||q2|/r^2
k=8.99*10^9 Nm^2/C^2
1 nC = 10^-9 C

3. The attempt at a solution
a. equation setup: .04 N = 8.99*10^9 * (q1)(q2)/(.23m)^2
I got 1.2x10^-6 for q1 & q2 as my answer, but that's not correct. Was my setup wrong or was it just a simple alegbraic error?

b. I honesty have no clue how to solve this problem. Hopefully someone could give me a little insight on how to setup this problem.

c. This problem seemed easy enough and straight forward, but after several attempts I still get the wrong answer. Here's my equation setup: F = 8.99*10^9 * (-1.3*10^-3)(1.5*10^-8)/(.073m)^2 I got 3.3*10^-4 but apparently that's not the correct answer.

d. I honestly don't know have to solve this one either. :(

2. Sep 6, 2009

### ideasrule

It's a simple algebraic error. Recalculate, and remember that q is negative.

It's asking for the spring constant, so the equation F=-kx should pop into mind. Since k=-F/x, what's F? What's x?

You've found the force between the particles, but the question is asking for the electric field. E=kq/r2 for each particle.

Can you find the horizontal component of the electric field? The vertical component? (Hint: Use E=kq/r2.)

3. Sep 6, 2009

### Anthem26

ok, i tried to work out a and b, but still no luck. here's my work:

(a) Given: r=.23m, q1=q2, f=.04N

Equations needed:F=k((q1q2)/r^2), where k is approx 8.9910^9 Nm^2/C^2

Solution: Since q1=q2, we can substitute q2 for q1 in the equation

F=k((q1q1)/r^2)

F=k*((q1^2)/r^2)

F/k=(q1^2)/r^2

r^2*(F/k)=(q1^2)

q1=q2=sqrt(r^2*(F/k))

plug in the given values....sqrt(.23^2(.04/(8.9910^9)))

i get approx -4.85*10^-7 C

*EDIT ok i managed to get this right, I just had to rewrite 4.85*10^7 to .000000485
(b) Using the E=kq/r2 provided. I plugged in:
E1 = 8.99*10^9 * (13*10^-6)/(.073m)^2
= 2.2*10^7

E2 = 8.99*10^9 * (15*10^-6)/(.073m)^2
= 2.5*10^7

E = 2.5*10^7 - 2.2*10^7 = 3*10^6

What did I do wrong?

Last edited: Sep 6, 2009