# Do i use coulomb's law?

1. Dec 18, 2003

### mustang

Problem 11.
Find the magnitude electric field at a point midway between two charges of 34.4*10^9 C and 78.6*10^-9 C separated by a distance of 58.2 cm. Answer in N/C.
Note: Do i use coulomb's law? If so, when i multiply the constant to the quiotent is that my answer?

Problem 14.
An electron moving through an electric field experiences an acceleration of 6*10^3m/S^2.
a. Find the magnitude of the electric force acting on the electron. Answer in N.
Note: Do i use F=mass*acceleration?
b. What is the magnitude of the electric field strength? Answer in N/C.
What formula do I use?

2. Dec 18, 2003

### himanshu121

E=qF will help u
F=ma will do

3. Dec 19, 2003

### ShawnD

Re: physics

Find the field from 1 then the field from the other then subtract.

field from first charge:
$$F = \frac{k q_1 q_2}{d^2}$$

One of the charges isn't there so just divide it out.
$$\frac{F}{q_1} = \frac{k q_2}{d^2}$$

$$\frac{F}{q_1} = \frac{(9x10^9)(34.4x10^-^9)}{0.291^2}$$

$$\frac{F}{q_1} = 3656$$ N/C

field from second charge:
$$\frac{F}{q_1} = \frac{k q_2}{d^2}$$

$$\frac{F}{q_1} = \frac{(9x10^9)(78.6x10^-^9)}{0.291^2}$$

$$\frac{F}{q_1} = 8354$$ N/C

field at that point:
8354 - 3656 = 4698 N/C

For question A:
You know the formula $$F = ma$$. You know the mass of an electron and its rate of acceleration. Sub in to find the force.

For question B:
The force is given by the field strength x charge. For field I'll just put L since I don't know what it should be.
F = Lq
L = F/q

You solved the force in part A and you know the charge of an electron.

Last edited: Dec 19, 2003
4. Dec 19, 2003

### mustang

Sorry

I rechecked and for problem 11 it was 34.4*10^-9C.

5. Dec 19, 2003

### himanshu121

so the answer will be 4698 * 10^-9

6. Dec 19, 2003

### mustang

Question for problem 14.

I had a=6*10^3 and m=9.109*10^-31. I subsituted those values for ma in F=ma and got 5.4654*10^-27. When i posted the answer I got it wrong. What did I do wrong?