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
Let's see if I can explain why part b0 can't be solved.
Let's assume the charges are 2m apart so r is always 1. Now suppose I make q=1e-3 and q' = 0
E = kq/r^2 for each and half way the field is 9e6. So for part b) E = 18e6.
Now suppose I start with q = 2e-3 and q' = 1e-3. Again half way E = 9e6 as proscribed. But now for part b) E = 36e6. There's no one answer. In fact if I'm not mistaken there are infinite solutions.
Bob? That sound about right?
To determine the resultant electric field at point O, we need to consider the electric field created by each charge individually and then add them vectorially.
a) When q' = q:
In this case, both charges are equal in magnitude. We can assume that the charges are positive for simplicity.
1. Electric field created by charge q at point O:
Given that the electric field strength E = 9 × 10^6 V/m, this field is created by charge q at point O. We know that electric field due to a point charge is given by the Coulomb's Law:
E = k * (q / r^2)
Where k is the electrostatic constant (8.99 × 10^9 Nm^2/C^2), q is the charge magnitude, and r is the distance between the charge and point O.
From the given information, E = 9 × 10^6 V/m. Plugging these values into the equation, we get:
9 × 10^6 = k * (q / r^2)
2. Electric field created by charge q' at point O:
Since q' has the same magnitude as q, it also creates an electric field at point O. By analogy to the previous step, we can write:
9 × 10^6 = k * (q' / r^2)
But q' = q, so:
9 × 10^6 = k * (q / r^2)
Since both equations are identical, the resultant electric field at point O when q' = q is also E = 9 × 10^6 V/m.
b) When q' = -q:
In this case, the charges have equal magnitudes but opposite signs. We'll consider q as positive and q' as negative.
1. Electric field created by charge q at point O:
Using the equation mentioned earlier, we can write:
9 × 10^6 = k * (q / r^2)
2. Electric field created by charge q' at point O:
Since q' = -q (opposite sign), the equation becomes:
9 × 10^6 = k * (-q / r^2)
To determine the resultant electric field, we need to add the vector components of the two fields. Since both fields point in the same direction (towards point O), we can simply add them algebraically:
Resultant electric field at O = E1 + E2 = 9 × 10^6 - 9 × 10^6 = 0 V/m
Therefore, the resultant electric field at point O when q' = -q is zero (0 V/m).