Two point charges q and q' are placed respectively at two points A and B. Let O be the midpoint of [AB].
Charge q placed at A creates at O an electric field of strength E=9×10^6 V/m.
Determine the resultant electric field at O when:
a) q'=q
b) q'=-q
a) zero obviously
b) can't really tell. just saying E = 9e6 half way tells you nothing about the individual charges. Is one positive and the other negative? Are both positive?
To determine the resultant electric field at point O, we need to consider the electric fields created by each charge individually and then add them vectorially.
a) When q' = q:
Since q' is equal in magnitude to q, the electric field created by q' at point O will also have the same magnitude and direction as the electric field created by q at O. So, the resultant electric field at O is equal to the sum of the electric fields created by q and q' at O.
To find the magnitude of the resultant electric field, we can use the formula for the electric field due to a point charge:
E = k * |Q| / r^2
Given that the electric field due to charge q at O is 9×10^6 V/m, we can use this equation to find the magnitude of q:
9×10^6 = k * |q| / r^2
Now, since O is the midpoint of AB, we can assume that AB has a length of '2r' (to keep things general). So, r = AB/2 = r.
Therefore, the equation becomes:
9×10^6 = k * |q| / (r^2/4)
Simplifying the equation, we get:
|q| = (9×10^6 * r^2) / k
Now, we know the magnitude of q. To find the resultant electric field at O, we can use the formula:
E_res = |E_q| + |E_q'|
Given that |E_q| = |E_q'| = 9×10^6 V/m, the resultant electric field at point O will be:
E_res = 9×10^6 V/m + 9×10^6 V/m = 18×10^6 V/m
So, the resultant electric field at O is 18×10^6 V/m when q' = q.
b) When q' = -q:
In this case, q' has the same magnitude as q but an opposite sign. This means that the electric field created by q' at O will have the same magnitude as the electric field created by q at O but in the opposite direction.
So, the resultant electric field at O is equal to the difference between the magnitudes of the electric fields created by q and q' at O, taking into account their opposite directions.
Using the same formulas as before, the magnitude of the electric field due to q is 9×10^6 V/m.
The magnitude of the electric field due to q' will be:
|E_q'| = k * |-q| / r^2 = k * |q| / r^2 = 9×10^6 V/m
Now, we can calculate the resultant electric field:
E_res = |E_q| - |E_q'|
E_res = 9×10^6 V/m - 9×10^6 V/m = 0 V/m
So, the resultant electric field at O is 0 V/m when q' = -q.