# You construct a rectangle that is 6cm by 8cm

1. Oct 3, 2007

### Xaspire88

The E-field at the center of 4 charges

You construct a rectangle that is 6cm by 8cm, and place the following four charges at each corner: 2mC, -4mC, 2mC, -3mC. The 2mC starts at the top left corner and then each is placed around the rectangle in order in a clockwise manner.

A) What is the E-field at the center of the rectangle?

B) What is the Potential at the center of the rectangle?

The only way i could think of to go about solving this was to separate the rectangle in to four triangles, and then solve for each of their Fnetx and Fnety to determine the Net force and direction.

F= (Kq1q2)/(r^2)

after finding the net force i figured i could use Fnet= qE E- being the E-field
but in order to do this i guess i would have to place some sort of charge at the middle of the rectangle to solve for E. i suppose i could use an electron. Would this work? or is there a simpler way i am missing?

for the next part i suppose id have to find out what U is for the system. V being potential what i need,
V=Ue/q Ue=(Kq1q2)/r

let me know what you think.

Last edited: Oct 3, 2007
2. Oct 3, 2007

### Astronuc

Staff Emeritus
Ei= (Kqi)/(ri^2), where qi represents the ith charge and ri the distance from the ith charge to the point of interest. It is a vector field.

3. Oct 3, 2007

### Xaspire88

i understand that its a vector.. but i have 4 different charges so how do i relate them all together to arrive at the resulting E-field?

4. Oct 4, 2007

### Xaspire88

Hmmm..

these are the components of the e field

E2mc= K(2x10^-3)/(5x10^-2)^2= 7.192 x 10^9 N/C

E-4mc= K(4x10^-3)/(5x10^-2)^2= 1.4384 x 10^10 N/C

E2mc= K(2x10^-3)/(5x10^-2)^2= 7.192 x 10^9 N/C

E-3mc= K(-3x10^-3)/(5x10^-2)^2= 1.0788 x 10^10 N/C

it seems the the components of the e field produced by the 2mC charges cancel and the resulting e field would be that of the -4mC charge and the -3mC charge.. they are acting in opposite directions... would i subtract them from each other. would the resulting direction be that of the greater component? being the -4mC charge? so the e field would be 3.596 x 10^9 N/C to the top right direction at an angle of 53.1 degrees?

5. Oct 4, 2007

### Xaspire88

also for the second part, finding the potential at the center, potential is a scalar quantity so v= sum of all potentials
V= kq1/r1 + Kq2/r2 ...... and so on

but since i have positive and negative charges do i subtract them from each other or add them all together as positives?

V= K(2mC)/5cm + or - K(3mC)/5cm?

6. Oct 4, 2007

### Xaspire88

Any thoughts? anyone?

7. Oct 4, 2007

### learningphysics

For part b), yes add up the kq/r for each one... you need to keep the signs... so it comes out to k*(-3mC)/5cm. of course convert the units and everything to get your final answer...

For part a), the fields due to the two 2mC charges cancel... because the fields are equal in magnitude but opposite in direction...

so you're left with these 2:

E-4mc= K(4x10^-3)/(5x10^-2)^2= 1.4384 x 10^10 N/C

E-3mc= K(-3x10^-3)/(5x10^-2)^2= 1.0788 x 10^10 N/C

these look right to me as the magnitudes of the two fields... but it would be best to get them in vector form... ie in the form $$E_x\vec{i} + E_y\vec{j}$$.

now what is the x-component of the E-4mc field... take right and up as positive, down and left as negative... and what is the y-component?

same way get the x-component and y-component of the E-3mc field...

careful about signs and directions here...

so once you know the x and y components... you add the x-components... add the y-components... to get the net field at the center.