two equal positive charges q are on the y
axis,one at y=+a and the other at y=-a.
a)show that the electric field on the x axis with < Ex=2kqx/(x^2+a^2)^3/2 > .
b)show that near the origin,when x is much smaller than a,E,is approximately 2kqx/a^3.
To solve this problem, we can use the principles of electrostatics and the concept of superposition. We'll break down the problem into parts.
a) To find the electric field on the x-axis at a point with coordinates (x, 0), we need to calculate the contributions from each charge individually and then add them up.
Let's denote the charge at y = +a as q1 and the charge at y = -a as q2. The electric field due to each charge can be calculated using Coulomb's law:
For q1 at (0, a):
E1x = k * q * (x - 0) / [(x - 0)^2 + a^2]^(3/2)
= k * q * x / (x^2 + a^2)^(3/2)
For q2 at (0, -a):
E2x = k * q * (x - 0) / [(x - 0)^2 + a^2]^(3/2)
= -k * q * x / (x^2 + a^2)^(3/2)
The total electric field on the x-axis is the sum of the electric fields due to q1 and q2:
Ex = E1x + E2x
= k * q * x / (x^2 + a^2)^(3/2) - k * q * x / (x^2 + a^2)^(3/2)
= 2kqx / (x^2 + a^2)^(3/2)
So, we have shown that the electric field on the x-axis, Ex, is given by Ex = 2kqx / (x^2 + a^2)^(3/2).
b) To show that near the origin, when x is much smaller than a, Ex is approximately 2kqx / a^3, we can make the approximation x << a and simplify the expression for Ex.
Using the binomial approximation formula (1 + ε)^n ≈ 1 + nε, where ε is a small number, we can simplify the denominator:
(x^2 + a^2)^(3/2) ≈ (a^2)^(3/2) = a^3
Substituting this approximation back into the equation for Ex, we get:
Ex ≈ 2kqx / (a^3)
= 2kqx / a^3
Therefore, we have shown that near the origin, when x is much smaller than a, the electric field Ex is approximately 2kqx / a^3.