Check out some similar questions!

Rubentheran · Feb 22, 2011, 08:51 AM

Three charges are placed at three corners of a square as shown in Figure 5. Find the electric field strength at point A.

Given answer is 8.23x10^5 N/C and makes an angle 45 degree below Ã¢â‚¬â€œx-axis.

I can't get the answer.First I start the formula E=k(q1xq2)/r^2 and x and y component then I don't know how and what formula to use but I got try my best to do this question but I still unable to find solution.Can someone guide me with solution nicely.Thanks in advance

ebaines · Feb 22, 2011, 11:09 AM

First, the magnitude of the electric field is given by:

$<br /> |\vec E| = k \frac q {r^2}<br />$

And for point charges like this the direction of the electric field is away from the charge if the charge is positive, and toward the charge if it is negative. So first calculate the x and y components of the electric field at point A due to each of the three charges, then add the x and y components togther, and lastly combine these values using vector addition.

I'll help you get started, For the -4\ $micro$ C charge the magnitude of the electric field at A is:

$<br /> |E| = k \frac q {r^2} = 8.99 \times 10^6 \frac {Nm^2}{C^2} \ \times \ \frac {4 \times 10^{-6}C} {(0.2m)^2} = 449500 \frac N C<br />$

The x and y components of this field are:

E_x = E sin(45) = 318000 N/C
E_y = E cos(45) = 318000 N/C

Note these are positive values because the direction is up and towards the right (in the direction toward the charge).

Now repeat for the other two charges, and add the three elecric field vectors together. Can you take it from here?