A charge-Q is located at x=-l/2 and a charge +Q is located at x=l/2 . Thus the separation between the two charges is l .
The total electric field on the x axis can be written as
E(x)=E(x)d^
(a) What is the direction d^ of the total electric field at any point on the x axis where x>l/2 ?
(b)What is E(X) as a function of x for x>l/2? Express your answer in terms of, if necessary, Q ,l , x and the constant epsilon_0 (if needed, enter pi for π , epsilon_0 for epsilon_0 ).
unanswered
(c) Consider now the limit where x>>l , so that
(1±l/2x)^-2~1干l/x
Express, in this limit,E(x) in terms of, if necessary, p , x and epsilon_0 . The quantity P=Ql is called the dipole moment.
To find the direction of the total electric field at any point on the x-axis where x > l/2, we can use the principle of superposition. Since both charges have the same magnitude Q but opposite signs, the electric fields they produce will also have opposite directions.
(a) At any point on the x-axis where x > l/2, the electric field from the negative charge at x = -l/2 will point towards the left (negative x-direction), while the electric field from the positive charge at x = l/2 will point towards the right (positive x-direction). Therefore, the total electric field at any point on the x-axis where x > l/2 will be the sum of these two electric fields, resulting in a net electric field pointing towards the positive x-direction.
Now let's move on to part (b) and find E(x) as a function of x for x > l/2. To do this, we can use the formula for the electric field due to a point charge:
E = k * Q / r^2
where E is the electric field, k is Coulomb's constant (k = 1 / (4πε₀) with ε₀ being the permittivity of free space), Q is the charge, and r is the distance from the charge.
For any point on the x-axis where x > l/2, the distances from the negative charge at x = -l/2 and the positive charge at x = l/2 can be calculated as follows:
Distance from the negative charge:
r₁ = √((-l/2 - x)^2) = √(l²/4 + x² - lx)
Distance from the positive charge:
r₂ = √((l/2 - x)^2) = √(l²/4 + x² - lx)
Using the electric field equation, the electric fields at any point on the x-axis where x > l/2 due to the negative charge and the positive charge respectively can be written as:
E₁ = k * Q / r₁²
E₂ = k * Q / r₂²
Since the electric field is a vector quantity, the total electric field E(x) at any point where x > l/2 will be the vector sum of E₁ and E₂:
E(x) = E₁ + E₂
Substituting the values of r₁ and r₂ and using the expression for k, the equation becomes:
E(x) = (1 / (4πε₀)) * (Q / (l²/4 + x² - lx)) + (1 / (4πε₀)) * (Q / (l²/4 + x² - lx))
Simplifying the expression further, we get:
E(x) = (Q / (4πε₀)) * (1 / (l²/4 + x² - lx))
This is the expression for E(x) as a function of x for x > l/2 in terms of Q, l, x, and ε₀.
Moving on to part (c), let's consider the limit where x >> l. In this limit, the terms l²/4 and lx become negligible compared to x². Using this approximation, the expression for E(x) simplifies as follows:
E(x) = (Q / (4πε₀)) * (1 / (x²))
This expression is now in terms of the dipole moment P = Q * l, x, and ε₀.
Note: It is important to remember that these derivations and approximations are based on the given assumptions and conditions.