# Need help on homework

1. Aug 31, 2009

### mustang1988

1. An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 120 m/s2. What are the magnitude and direction of the electric field?

2. f=ma

3. F=(9.11e-31)(120m/s^2)=1.1e-28

2.Two small nonconducting spheres have a total charge of 92.4 µC
a) when placed 1.00 m apart, the force each exerts on the other is 12.0 N and is repulsive. What is the charge on each? Smaller and Larger
b)What if the force were attractive?

If anyone could help me that would be great, i just started physics 2 a year after physics 1 so i am having some difficulties. I thought i did the first one right but the computer says i have the wrong answer, and i dont really know where to start on the second. Thanks

2. Aug 31, 2009

### rock.freak667

So you have the force, now what is the electric field defined as?

the electric force between two point charges, Q1 and Q2, separated by a distance r is given by:

$$F=\frac{Q_1 Q_2}{4 \pi \epsilon r^2}$$

You will need this equation and what is in bold

3. Aug 31, 2009

### rl.bhat

In the problem 1, force on the electron F = E*q = m*a where E is th electric field, q is the charge on electron. Now find E.
In the problem 2, q1 = q and q2 = 92.4 μC - q, if q1 and q1 are positive.
What should the charges for repulsive force?

4. Sep 1, 2009

### mustang1988

For Number 1 would i just use F=QE where F=1.1e-28, Q=1.6e-19 then solve for E?

5. Sep 1, 2009

### mustang1988

And for Number 2 wont i still have two unknowns, F and q? Sorry im just really lost on this one. thanks

6. Sep 2, 2009

### rl.bhat

From the problem it implies that the attractive or the repulsive force has the same magnitude i.e. 12 N.

7. Sep 2, 2009

### mustang1988

Thanks for the help i got the right answers. Is there an easier way then doing the quadradic equation though? It was a pain carrying all the numbers through

8. Sep 2, 2009

### rl.bhat

Since the sum of the charges is given, you cant avoid the quadratic equation.

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