Two charges are placed on the x axis. One of the charges (q1 = +8.59C) is at x1 = +3.00 cm and the other (q2 = -25.6C) is at x2 = +9.00 cm. Find the net electric field (magnitude and direction given as a plus or minus sign) at (a) x = 0 cm and (b) x = +6.00 cm.
Well, it looks like we've got some electrifying charges on the loose here! Let's calculate the net electric field at x = 0 cm and x = +6.00 cm.
To find the net electric field at a point, we need to calculate the electric field due to each charge and then add them up (just like counting your sheep before going to sleep, but with more numbers).
First, let's tackle the electric field at x = 0 cm. Taking into account the distances and charges, we can determine the electric field due to q1 and q2.
Electric field due to q1:
E1 = k * (q1 / r1^2)
Electric field due to q2:
E2 = k * (q2 / r2^2)
Where k is Coulomb's constant (k ≈ 9 x 10^9 Nm^2/C^2), r1 is the distance between q1 and the point (0 cm in this case), and r2 is the distance between q2 and the point.
Now, if both charges have the same magnitude (without the sign), let's call it q, we can simplify the equation:
E1 = k * (q / r1^2)
E2 = k * (q / r2^2)
Now, since q1 is positive, its electric field points away from it (+), and since q2 is negative, its electric field points toward it (-). So when we add both electric fields, we subtract their magnitudes.
Magnitude of the net electric field at x = 0 cm:
|E_net(0 cm)| = |E1 - E2|
Now, let's repeat the same procedure to find the net electric field at x = +6.00 cm. We just need to substitute the distances (r1 and r2) accordingly.
Magnitude of the net electric field at x = +6.00 cm:
|E_net(+6.00 cm)| = |E1 - E2|
And there you have it! Just plug in the numbers and calculate the magnitudes of the net electric fields. Remember to pay close attention to the signs of the charges and whether they attract or repel each other. Have a shocking good time with your calculations!