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
Thanks
To determine the resultant electric field at point O, we need to consider the contribution of electric fields created by both charges q and q' separately.
a) When q' = q:
In this case, since the charges are the same, the electric fields created by q and q' will have the same magnitude and direction.
Let's denote the electric field created by q at point O as EO(q) and the electric field created by q' at point O as EO(q').
Since the electric field strength E = 9×10^6 V/m is given for EO(q), it means that the magnitudes of EO(q) and EO(q') will be equal to that value.
Therefore, the resultant electric field at O when q' = q is:
EO(resultant) = EO(q) + EO(q') = 9×10^6 V/m + 9×10^6 V/m = 18×10^6 V/m = 1.8×10^7 V/m
So, the resultant electric field at O when q' = q is 1.8×10^7 V/m.
b) When q' = -q:
In this case, since the charges have opposite signs, the electric fields created by q and q' will have opposite directions.
Let's denote the electric field created by q at point O as EO(q) and the electric field created by q' at point O as EO(q').
Since the electric field strength E = 9×10^6 V/m is given for EO(q), the magnitude of EO(q') will also be 9×10^6 V/m.
However, since q' = -q, the direction of EO(q') will be opposite to that of EO(q).
Therefore, the resultant electric field at O when q' = -q is:
EO(resultant) = EO(q) - EO(q') = 9×10^6 V/m - 9×10^6 V/m = 0 V/m
So, the resultant electric field at O when q' = -q is 0 V/m.
To determine the resultant electric field at point O, we need to calculate the individual electric fields created by each charge q and q', and then add them together using vector addition. Let's start with the given information.
a) q' = q
Since q' is equal in magnitude to q, both charges at points A and B will create electric fields of the same magnitude at point O. The resultant electric field at O will be the sum of the two electric fields:
E_resultant = E_q + E_q'
b) q' = -q
In this case, the charges at points A and B have opposite magnitudes. Since the electric field is directly proportional to the magnitude of the charge, the electric field created by q' at point O will be in the opposite direction as the electric field created by q. The resultant electric field at O will again be the sum of the two electric fields:
E_resultant = E_q + E_q'
Now, let's calculate the electric fields created by each charge and find the resultant electric field in each case.
Given:
E = 9 × 10^6 V/m
a) q' = q
Since both charges have the same magnitude, the electric fields created by each charge at O will be the same:
E_q = E
E_q' = E
Therefore, the resultant electric field at O is twice the individual electric field:
E_resultant = E_q + E_q'
E_resultant = E + E
E_resultant = 2E
Substituting the given value, we get:
E_resultant = 2(9 × 10^6 V/m)
E_resultant = 18 × 10^6 V/m
E_resultant = 1.8 × 10^7 V/m
b) q' = -q
Since the charges have opposite magnitudes, the electric fields created by each charge at O will have equal magnitudes but opposite directions:
E_q = E
E_q' = -E
Therefore, the resultant electric field at O is the difference between the two electric fields:
E_resultant = E_q + E_q'
E_resultant = E + (-E)
E_resultant = 0
So, the resultant electric field at O when q' = -q is zero.
To summarize:
a) When q' = q, the resultant electric field at O is 1.8 × 10^7 V/m.
b) When q' = -q, the resultant electric field at O is 0.