A 25mC charge is placed at the end of a meter stick and a 4.omC charge is placed at the 50cm mark.At what position on the line joining the two charges is the electric field zero?
The field will be zero where the distance to the distances are in the ratio sqrt(25/4)= 5/2, with the closest distance to the larger charge.
That will make the Coulombb forces equal and opposite. Since the sum of the two distances is 50 m, the location is 2/7 of the way from the end to the middle, at the 14.29 cm mark. That will be 35.71 cm from the 4.0 mC charge.
The ratio of the two distances is 2.5, as required.
To determine the position on the line joining the two charges where the electric field is zero, we can use the concept of the electric field due to point charges.
The electric field at any point in space is determined by the vector sum of the electric fields produced by each charge at that point.
In this case, we have two point charges: a +25mC charge at one end of the meter stick and a +4.0mC charge at the 50cm mark.
To find the position where the electric field is zero, we need to determine the point where the electric fields of both charges cancel out each other. This can be done by setting up an equation based on the principle that the vector sum of the electric fields is zero.
Let's assume that the position where the electric field is zero is at 'x' cm from the +25mC charge. The position of the +4.0mC charge is at 50cm.
The electric field due to the +25mC charge at position x is given by:
E1 = k * q1 / r1^2
The electric field due to the +4.0mC charge at position 50cm is given by:
E2 = k * q2 / r2^2
Since the electric fields target opposite directions, to cancel each other, they must have equal magnitude and opposite sign:
E1 = -E2
Setting up the equation:
k * q1 / r1^2 = - k * q2 / r2^2
Plugging in the given values:
(9 * 10^9 Nm^2/C^2) * (25 * 10^-3 C) / (x * 10^-2 m)^2 = - (9 * 10^9 Nm^2/C^2) * (4 * 10^-3 C) / (50 * 10^-2 m)^2
Simplifying the equation:
25 / x^2 = -4 / 5^2
Cross-multiplying:
25 * 5^2 = -4 * x^2
Solving for x:
25 * 25 = 4 * x^2
625 = 4 * x^2
x^2 = 625 / 4
x^2 = 156.25
x ≈ 12.5 cm
Therefore, the position on the line joining the two charges where the electric field is zero is approximately 12.5 cm from the +25mC charge.