Multiple choice (show your work if any):
1) Two point charges q1=-10^-5 C and q2=-9×10^-5 C are placed respectively at two points A and B 40 cm apart. The electric field is null at a point C of [AB] such that :
a) AC=50cm b) AC=10cm c) AC=20cm d) point C doesn't exist
2) Same given. The electric field is null at a point C outside [AB] such that :
a) AC=50cm b) AC=30cm c) AC=20cm d) point C does not exist
Thanks.
To find the point where the electric field is null, we need to consider the superposition principle for electric fields created by the charges q1 and q2.
1) For the case where the electric field is null at a point C on the line segment [AB]:
Let's assume that the distance from point A to point C is x cm. Therefore, the distance from point B to point C would be (40 - x) cm.
According to the superposition principle, the electric field at point C can be calculated as follows:
E = k * (q1 / r1^2) + k * (q2 / r2^2)
where:
E is the electric field
k is the electrostatic constant (9 × 10^9 N m^2 / C^2)
q1 and q2 are the magnitudes of the charges
r1 and r2 are the distances from the charges to point C
Since we know that the electric field is null at point C, we can set E = 0 and solve for x.
0 = k * (q1 / r1^2) + k * (q2 / r2^2)
Simplifying and rearranging the equation:
(q1 / r1^2) = - (q2 / r2^2)
Substituting the given values:
(-10^-5 C) / (a^2) = (-9×10^-5 C) / ((40-a)^2)
Cross-multiplying and simplifying:
-10^5 (40 - a)^2 = -9 × 10^5 a^2
Expanding and simplifying:
1600 - 80a + a^2 = 9a^2
8a^2 + 80a - 1600 = 0
Dividing the equation by 8:
a^2 + 10a - 200 = 0
Factorizing the quadratic equation:
(a + 20)(a - 10) = 0
So, a = -20 or a = 10
Since distance cannot be negative, the valid solution is a = 10 cm.
Therefore, AC = 10 cm (option b) is the correct answer.
2) For the case where the electric field is null at a point C outside the line segment [AB]:
Using the same approach as before, we can set up the equation:
(q1 / r1^2) = - (q2 / r2^2)
Substituting the given values:
(-10^-5 C) / (a^2) = (-9×10^-5 C) / ((40+a)^2)
Cross-multiplying and simplifying:
-10^5 (40 + a)^2 = -9 × 10^5 a^2
Expanding and simplifying:
1600 + 80a + a^2 = 9a^2
8a^2 - 80a - 1600 = 0
Dividing the equation by 8:
a^2 - 10a - 200 = 0
Factorizing the quadratic equation:
(a - 20)(a + 10) = 0
So, a = 20 or a = -10
Since distance cannot be negative, the valid solution is a = 20 cm.
Therefore, AC = 20 cm (option c) is the correct answer.
To determine the answer to these multiple-choice questions, we need to analyze the electric field due to the two point charges and find the conditions for the electric field to be null at point C.
1) The electric field due to a point charge is given by the formula: E = k*q/r^2, where k is the electrostatic constant (k ≈ 9 x 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge.
At point C, if the electric field is null, that means the electric field due to q1 and q2 cancels each other out. This happens when the magnitudes of the electric fields due to both charges are the same but have opposite directions.
Let's calculate the electric field due to each charge at point C:
For q1: E1 = k*q1/r1^2, where r1 represents the distance between q1 and point C.
For q2: E2 = k*q2/r2^2, where r2 represents the distance between q2 and point C.
Since the electric fields must be equal in magnitude but opposite in direction, we can write:
|E1| = |E2|
Now, let's substitute the given values into these equations:
For q1: E1 = (9 x 10^9 Nm^2/C^2) * (-10^-5 C) / r1^2
For q2: E2 = (9 x 10^9 Nm^2/C^2) * (-9 x 10^-5 C) / r2^2
To cancel out the electrostatic constant, we can simplify the equation:
|E1| = |E2|
|-10^-5 / r1^2| = |-9 x 10^-5 / r2^2|
To simplify further, we can remove the negative signs:
10^-5 / r1^2 = 9 x 10^-5 / r2^2
Now, we need to solve for the distance ratio between r1 and r2 (r1/r2):
10^-5 * r2^2 = 9 x 10^-5 * r1^2
r2^2 / r1^2 = 9
sqrt(r2^2 / r1^2) = sqrt(9)
|r2/r1| = 3
Since distances cannot be negative, r2/r1 = 3. Now let's apply this to find the possible answers:
AC = r1 + r2 = r1 + 3*r1 = 4*r1
Since we know that AB = 40 cm, we can write that AC = 4*r1 = 40 cm.
From this, we conclude that AC = 10 cm (Option b) is the correct answer.
2) In this scenario, the point C is outside the line segment AB. We need to find the conditions for the electric field to be null at point C.
Using the same analysis as in the previous question, we know that the magnitudes of the electric fields due to both charges must be equal, but they have opposite directions.
Let's set up the equations for the electric fields at point C:
For q1: E1 = k*q1/r1^2
For q2: E2 = k*q2/r2^2
The condition for the electric field to be null is given by |E1| = |E2|. Applying this condition:
|-10^-5 / r1^2 | = | -9 x 10^-5 / r2^2|
To simplify further, we can remove the negative signs:
10^-5 / r1^2 = 9 x 10^-5 / r2^2
Now, solving for r2/r1:
10^-5 * r2^2 = 9 x 10^-5 * r1^2
r2^2 / r1^2 = 9
|r2/r1| = 3
From this, we conclude that point C is located at a distance 3 times the distance of B from A (AC = 3*AB).
Since AB = 40 cm, we find AC = 3*40 cm = 120 cm.
Therefore, point C does not exist (Option d) is the correct answer.