# Physics problem

1. Dec 20, 2003

### mustang

Problem 16.
Three charges: +8.2uC 4.8 cm to the left of 4uC and a -2.3uC 2.1 cm to the right of the 4uC charge.
What is the electric field strength at a point 2.7cm to the left of the middle charge? In N/C.
Note" Is the answer 1.733218664*10^7 from the 8.99*10^9 (4*10^-6)/(0.027^2)=49327846.36 and (8.99*10^9)(8.2*10^-6)/(0.048^2)=31995659.72 which were subtracted to get that answer.

2. Dec 20, 2003

### gnome

You might want to look that over again.

For starters, there are 3 charges & you figured the field from only 2 (& got one of the distances wrong).

Draw a diagram first, & work from that.

3. Dec 20, 2003

### mustang

I drew a diagram and have 8.2 uC 2.1 cm away from the point that is 2.7 cm to the left of the middle charge. In addition should i multiply 8.99*10^9 to -2.3uC divided by 3.1 cm and subtract what is now three values to get my answer?

4. Dec 20, 2003

### gnome

First, where did you get 3.1 cm?

Second, make sure you keep the directions straight. The field from the positive 8.2 &mu;C charge is directed toward the right. What are the directions of the other two fields?
i.e.: same direction = add; opposite direction = subtract

And don't forget, you are dividing by the square of the distance.

5. Dec 20, 2003

### mustang

woops! From -2.3uC to 4 uC is 2.1cm and from 4 uC it is 2.7 cm to reach that point so the distance would be 4.8cm. So for -2.3uC I multiply 8.99*10^9 to -2.3uC divided by 4.8 cm or 0.048m?

6. Dec 20, 2003

### gnome

.0482

But think carefully about the directions. It's not just a question of the sign of the charge. You have to consider the relative positions of the particles.

7. Dec 20, 2003

### gnome

The field from the +8.2 &mu;C charge has the same direction at point P as the -2.3 &mu;C charge.

Do you see why?

8. Dec 20, 2003

### mustang

No, i don't see why 8.2uC and -2.3uC have the same direction. I got three values from 4*106-6 is 49327846.36, from 8.2*10^-6 is 167160997.7, and -2.3*10^-6 is 8974392.361. So would i add 167160997.7 to 8974392.361 and subtract that from 167160997.7/

9. Dec 20, 2003

### gnome

No.

The direction of the electric field at any point P is the same as the direction of the electrical force that would be experienced by a positive "test" charge placed at that point.

In this problem, if you placed a positive test charge at point P, it would be repelled by the positive 8.2 &mu;C charge (call that charge A) towards the right since they're both positive, so the component of the field from charge A at point P is directed toward the right.

But the negative 2.3 &mu;C charge (call it C) would ATTRACT a positive charge, so a positive test charge located at point P would be pulled to the right. Therefore, the field component produced by charge C at point P is also directed toward the right.

Therefore, the field components of charges A and C at point P are added, not subtracted.

On the other hand, what would charge B (the +4 &mu;C charge in the middle) do to a positive test charge at point P?

Get it?

10. Dec 20, 2003

### mustang

Are the three values I got from the charges right?

11. Dec 20, 2003

### gnome

Yes. Now you just have to figure out what to add & what to subtract.

12. Dec 20, 2003

### mustang

So gnome since you said that "The field from the +8.2 ìC charge has the same direction at point P as the -2.3 ìC charge." I would add
167160997.7 to 8974392.361 to get 176135390.1. From that I would subtract 49327846.36 and get 126807543.7, right?

13. Dec 20, 2003

### gnome

Yes, but do you understand why, or are you just taking my word for it?