Homework Help: Homework help - Electric forces

1. Dec 4, 2004

Lorax

Homework help -- Electric forces

"Two tiny water drops, with identical charges -1.00x10^16C, have a centre to centre spereration of 1.00cm. A) What is the magnitude of the electrostatic force acting between them? B)How many excess electrons are on each drop, giving it its charge imbalance?"

Could someone get me started and help me walk through this? I dont want the answer I just want to be able to know what im doing to get the answer. ^^

2. Dec 4, 2004

Tide

If the charges are evenly distributed over the surfaces of the drops then you can treat the charges as point charges located at the center of their respective drops. For the second question, you just divide the total charge by the charge carried by a single elementary charge.

3. Dec 4, 2004

fannemel

For the first question you use the formula for electric force, hint: it's the one that resembles newtons law of gravity. For the second one, consider what charge N number of electrons sum up to.

4. Dec 4, 2004

Lorax

umm is this right for the first question....

Fe=(8.99x10^9)(-1.0x10^16)(-1.0x10^16)/(1.0x10^-2)^2?
Fe= 8.99 x10^47

Is that right and if it is what did you mean fannemel when you were helping me with the second question. I am confused thanks.

5. Dec 4, 2004

Lorax

While someone helps me out with that last question I will add my other Hw questions on here in hopes they too can be helped with.

85) The charges and coordinates of two charged particles held in a fixed xy plane are q1= -4.0uC, x1= 3.5cm, y1=0.5cm and q2= -4.0uC, x2= -2cm y2= -1.5cm.

A) Fine tbe magnitude and direction of the electrostatic force on q2.
B) Where would you locate a third charge q3 = +4.0uC such that the net electrostatic force is zero?

I'm stumped on this one and then theres....

109) The Electric potential difference between the ground and a cloud in a particular thunderstaorm is 1.2x10^9 V. What is the magnitude of the charge in the electrical potential energy (in multiples of the electron vold) of an electron that moves between the ground and cloud?

Do I use the equation |E|=v\d? using |E| as 1.2x10^9 and then d being 0 since it travels back and forth? I'm stumped with this aswell.

(I have one more but this hsould do for now)

6. Dec 4, 2004

Tide

Lorax,

(85) For this one you can once again use Coulomb's law to find the electric field produced by q1 and then mulitply by q2 to find the electrostatic force. To find the direction of the force remember that like charges repel and opposite charges attract and that the force acts along a line joining the two charges. I'll let you think some more about where you'd place a third charge to cancel out the electric field of the first one!

(109) To find the change in potential energy of the charge simply multiply the charge by the potential difference.

7. Dec 5, 2004

Lorax

Tide,

Alright for 85 I am a bit confused still as I thought columb's constant was Fe=kq1q2/R^2. I am uncertain as how to go about figuring out the electric fields. Do I put q1 and q2 together? or do it individually. And for that matter I am confused as to how to get R. R is the distance but I am uncertain how to do it for this question. A more in depth explanation would be grand.

And as for 109...

I did....

(1.2x10^9)(-1.6x10-19) and got 1.92 x 10 -10.

Is that right? Thanks in advance

Edit: Actually I got A) 8.99x10-19 and B)625
does that look right?

Last edited: Dec 5, 2004
8. Dec 7, 2004

fannemel

When i do the calculations myself i got
$$F_e = 8.99x10^{45}$$
but your first line is correct, probably just forgot to square the the distance when calculating the answer...

For the other question, we know that electrical charge is quantized and the smallest possible charge being the electron charge $$\mbox{q_e = 1.602x10^{19} C}$$. That means that all greater charges are just an integer multiplied with the electron charge. $$Q = nq_e$$

9. Dec 7, 2004

Tide

Lorax,

Sorry - I lost track of you!

The electric field at a point $\vec x_2$ produced by a point charge q1 at a point $\vec x_1$ is

$$\vec E = \frac {q_1}{4 \pi \epsilon_0} \frac {\vec x_2 - \vec x_1}{\left|\vec x_2 - \vec x_1 \right|^3}$$

and the force on a charge $q_2$ at point $\vec x_2$ is

$$\vec F = q_2 \ \vec E$$