Simple physics help

Hi, i have a physics problem i can't solve....

In one point A there is a positive charge Q1=50nC and in another point a negative charge Q2=-80 nC. AB=0,20m Determine the acceleration(size and direction) for an electron situated in the middle of AB.

I don't even know how to begin to solve this one

Quote:

Originally Posted by Jones

Hi, i have a physics problem i can't solve....

In one point A there is a positive charge Q1=50nC and in another point a negative charge Q2=-80 nC. AB=0,20m Detirmine the acceleration(size and direction) for an electron situated in the middle of AB.

I don't even know how to begin to solve this one

You know how to calculate the force on a charge due to another charge, right? (Coulomb's Law) Now recall that force is a vector. So we need to find the vector sum of the forces on the electron due to the charges at points A and B.

Define midpoint of AB to be the origin (ie. I'm placing the origin at the electron.) I am going to define a positive direction to the right. (Point A is at -0.10 m and point B is at 0.10 m.) NOTE: I am taking the absolute value of the charges in the force formula and figuring out the direction of the force on the electron by whether the electron is attracted or repelled by the charge.

The force on the electron by the charge at A is:
$F_{eA} = -\frac{kq_eQ_1}{r_{eA}^2} = -\frac{9 \times 10^9 \cdot 1.60 \times 10^{-19} \cdot 50 \times 10^{-6}}{0.10^2}$ N
(This force is in the negative direction since the two charges attract, indicated by the "-" sign out front.)

The force on the electron by the charge at B is:
$F_{eB} = -\frac{kq_eQ_2}{r_{eB}^2} = -\frac{9 \times 10^9 \cdot 1.60 \times 10^{-19} \cdot 80 \times 10^{-6}}{0.10^2}$ N
(This force is in the negative direction since the two charges repel, indicated by the "-" sign out front.)

Now just add the two forces. Finally, to get the acceleration of the electron, use F = ma. I would describe the acceleration as being in the direction of A or B. (Ans. The acceleration is in the direction of point A.)

-Dan

NOTE: In case you are using the "other" form of Coulomb's Law, use
$F = \frac{1}{4 \pi \epsilon _0 }\frac{q_1q_2}{r_{12}^2}$

The Coulomb constant, k is defined as $k = \frac{1}{4 \pi \epsilon _0} = 9 \times 10^9 \, \frac{N m^2}{C^2}$ .

Thank you very much for the help.

Funny i have never seen the definition of k before :o

Quote:

Originally Posted by Jones

Thank you very much for the help.

Funny i have never seen the definition of k before :o

My general experience is that engineering books tend to use $\epsilon _0$ and physics books use "k," but it really depends on the book. Obviously, the two approaches are equivalent.

-Dan