two point charges of 5 micro culomb and 20 micro culomb are separated by a distance of 2m find the point on the line joining them at which electric field intensity is zero
The point at which the electric field intensity is zero is 1m from the 5 micro culomb charge.
To find the point on the line of the two charges where the electric field intensity is zero, we can make use of the concept of the electric field due to point charges.
The electric field intensity at a point due to a point charge can be calculated using the equation:
E = k * (Q / r^2)
Where:
E is the electric field intensity,
k is the electrostatic constant (9 x 10^9 Nm^2/C^2),
Q is the charge,
and r is the distance from the charge.
Now let's calculate the electric field intensities due to the two charges at a general point (x) on the line joining them:
E1 = k * (5 micro C) / [(x - 1m)^2]
E2 = k * (20 micro C) / [(2m - x)^2]
Since we want the net electric field intensity to be zero, the magnitudes of E1 and E2 should be equal.
Therefore, we set E1 = E2:
k * (5 micro C) / [(x - 1m)^2] = k * (20 micro C) / [(2m - x)^2]
Simplifying the equation:
(1m - x)^2 = 4 * (2m - x)^2
Expanding and rearranging terms:
(x^2 - 2x + 1m^2) = 4 * (4m^2 - 4x + x^2)
Simplifying further:
x^2 - 2x + 1 = 16m^2 - 16x + 4x^2
Combining like terms:
3x^2 - 14x + 16m^2 - 1 = 0
This equation is a quadratic equation in the variable x. Solving this equation will give us the values of x at which the electric field intensity is zero.
Now, you can use the quadratic formula or factoring to solve this equation. The solutions for x will give us the points on the line where the electric field intensity is zero.
To find the point on the line joining two point charges where the electric field intensity is zero, we can use the concept of electric field vectors being additive.
Let's assume that the 5 μC charge (q1) is located at point A, and the 20 μC charge (q2) is located at point B. We need to find the point on the line AB where the electric field intensity is zero, which we'll call point P.
First, let's consider the direction of the electric field created by each charge individually:
1. For charge q1 (5 μC), the electric field created will be directed away from it (since it is a positive charge).
2. For charge q2 (20 μC), the electric field created will be directed towards it (since it is a negative charge).
At point P, the electric field created by q1 and q2 will cancel each other out, resulting in a net electric field intensity of zero.
Now, let's calculate the distances AP and PB. Since charge q1 is at A and charge q2 is at B, AP + PB should be equal to the total distance between them, which is 2 meters.
Let x be the distance AP, then the distance PB is 2 - x.
According to Coulomb's law, the electric field intensity (E) created by a point charge at a given distance is given by the equation:
E = k * q / r^2
Where k is the electrostatic constant (k = 9 * 10^9 Nm^2/C^2), q is the charge, and r is the distance.
1. Electric field created by q1 at point P (E1):
E1 = k * q1 / x^2
2. Electric field created by q2 at point P (E2):
E2 = k * q2 / (2 - x)^2
Since the net electric field intensity at point P is zero, E1 + E2 = 0:
k * q1 / x^2 + k * q2 / (2 - x)^2 = 0
Now, we can solve this equation to find the value of x, which will give us the position of point P on the line joining q1 and q2 where the electric field intensity is zero.
Simplifying the equation:
q1 / x^2 + q2 / (2 - x)^2 = 0
q1 * (2 - x)^2 + q2 * x^2 = 0
Solving this quadratic equation will give us the value/position of x, and as a result, point P, where the electric field intensity is zero.