What is the new force between two charges after they touch?

[six789]

Two oppositely charged objects exert a force of attraction of 8.0N on each other. What will be the new force of attraction if the objects are moved to a diustance four times their original distance of separation?

i did this..
F=kq1*q2/4r
8.0N=9.0x10^9 (1)(1)/4r
r=1.125x10^9m

how can you find the new force...HELP.. i don't know wat to do next here...
the right answer in my book is 0.50N.

 

[quasar987]

Um um, let's start this anew.

The electric force between two charged objects (provided they resemble dots enough) is given by

$$F = \frac{kq_1 q_2}{r^2}$$

Now, we are told that this force is of 8N for a particular distance $r$. In other words,

$$\frac{kq_1 q_2}{r^2} = 8 \ \ (*)$$

Now suppose we quadruple the distance between them. What is the electric force F', now that the distance between them is of $4r_{1,2}$? It is,

$$F' = \frac{kq_1 q_2}{(4r)^2} = \frac{kq_1 q_2}{16r^2}=\frac{1}{16} \left[ \frac{kq_1 q_2}{r^2} \right]$$

Now, making use of equation (*):

$$F' = \frac{1}{16} [ 8 ] = 0.50 N$$

 

[six789]

magnetic fields

how about this problem...
Two identical object have charges of +6.0µC and -2.0µC, respectively. When placed a distance d apart, their force of attraction is 2.0N. If the objects are touched together, then moved to d distance of separtion of 2d, what will be the new force between them?

i did this...
F=k*q1*q2/r^2
F is directly proportional to k*q1*q2/r^2
new force/2.0N = 9.0x10^9(6.0x10^-6C)\(2.0x10^-6C)/(2r)^2

is my work right? and i don't know the nexdt step here..
and the answer right answer is 0.17N (repulsive)

 

[quasar987]

This problem is based on the same principle as the first, but there is an important addition: the objects are put in contact before being distanced a distance 2d.

What happens when two charged object come in contact?

 

[asrodan]

Using the fact that the force between them is 2.0 N at distance d, you can find d.

What happens to the charges on the objects when you touch them together?

F2 = k*q3*q4/(2d)^2

 

[quasar987]

What happens to the charges on the objects when you touch them together?

F2 = k*q3*q4/(2d)^2

Right. The charges on each object will changes. But by how much?

Hint: Opposite charges attract each other and "cancel" when they come in contact with each other. Inversely, identical charges repulse each other. With this information, you can guess how the charges will rearange themselves on the two object when the two objects are in contact.

 

[six789]

so the equation is like this...
F is directly proportional to ((6.0x10^-6C)\(2.0x10^-6C)/(2r)^2)/(q1*q2/r^2)?

 

[quasar987]

WhaaAaT?

Start by answer the question: what is the new distribution of charge on the object due to the fact that they have been in touch?

 

[six789]

check this if this is right?

F/2N = 1.2x10^-11/4r^2

the new force i get is 6x10^-12N.. it is wrong, how can i get 0.17N? it's kinda hard to think about it...

 

[whozum]

If two charged conducting objects are touched, they will reach an equilibrium where the charge over both objects is equal.

 

[six789]

Um um, let's start this anew.

The electric force between two charged objects (provided they resemble dots enough) is given by

$$F = \frac{kq_1 q_2}{r^2}$$

Now, we are told that this force is of 8N for a particular distance $r$. In other words,

$$\frac{kq_1 q_2}{r^2} = 8 \ \ (*)$$

Now suppose we quadruple the distance between them. What is the electric force F', now that the distance between them is of $4r_{1,2}$? It is,

$$F' = \frac{kq_1 q_2}{(4r)^2} = \frac{kq_1 q_2}{16r^2}=\frac{1}{16} \left[ \frac{kq_1 q_2}{r^2} \right]$$

Now, making use of equation (*):

$$F' = \frac{1}{16} [ 8 ] = 0.50 N$$

whozum, i still don't get the concept behind this.. how can i apply if there is given charges??

 

[whozum]

Tell me exactly what you don't understand so we can work through

 

[six789]

whozum.. ok i get the process and solution you gave to me, but how can i apply the two charges... should i find the F first for the two charges and what should i do next??

 

[whozum]

Find the distance before they touch:

$$F = \frac{kq_1q_2}{r^2}$$

So

$$r = \sqrt{\frac{kq_1q_2}{F}$$

$$r =\sqrt{\frac{k(6uC)(-2uC)}{2N}$$ Take the magnitude of both charges to avoid complex numbers.

Multiply it by two to get the new distance.

Find the charge of equilibrium for both charges, that is, the charges they will have after touching. Someone left you a good hint about opposite and like charges. Use the force equation to find the new force.

 

[Doc Al]

how about this problem...
Two identical object have charges of +6.0µC and -2.0µC, respectively. When placed a distance d apart, their force of attraction is 2.0N. If the objects are touched together, then moved to d distance of separtion of 2d, what will be the new force between them?

i did this...
F=k*q1*q2/r^2
F is directly proportional to k*q1*q2/r^2
new force/2.0N = 9.0x10^9(6.0x10^-6C)\(2.0x10^-6C)/(2r)^2

is my work right? and i don't know the nexdt step here..
and the answer right answer is 0.17N (repulsive)

Let's call the force between the two charges before they are touched together $F_1$. (We know that force is 2.0N.)

Don't rush to calculate things, instead use ratios:

$$F_1 = k \frac{q_1 q_2}{r^2} \propto k \frac {(6) (-2)}{d^2} = - 12 k/d^2$$

This is an attractive force. Now quasar987 asked you twice to state what happens to the charges after the two object touch. That is the key! (Hint: Does the total charge change? How does the total charge distribute between the two objects when they touch?) Once you realize what the new charges are, calculate the new force $F_2$ just like I did above and compare it to $F_1$. (There is no need to calculate d, but you can if you want to.)