Question

To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positive charge Q distributed along its length. Find the magnitude of the electric field due to the wire at a point located a distance d from one end of the wire along the line extending from the wire.

Answer 1

To find the magnitude of the electric field due to a straight wire with a positive charge distributed along its length at a point located a distance d from one end of the wire, we can use the problem-solving strategy for continuous charge distribution problems.

Step-by-step explanation:

To find the magnitude of the electric field due to a straight wire with a positive charge distributed along its length at a point located a distance d from one end of the wire, we can use the problem-solving strategy for continuous charge distribution problems. We conceptually break the wire into differential pieces and calculate the differential field created by two symmetrically placed pieces of the wire using the symmetry of the setup. Finally, we integrate this differential field expression over the length of the wire to obtain the complete electric field expression.

Answer 2

E = k Q / [d(d+L)]

Step-by-step explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

λ = Q / L

If we derive from the length we have

λ = dq/dx ⇒ dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

dE = k dq / x²2

dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

E = k
`\int\limits^(d+L)_d {\lambda/x^(2)} \, dx`

We take out the constant magnitudes and perform the integral

E = k λ (-1/x)
`{(-1/x)}^(d+L) _(d)`

Evaluating

E = k λ [ 1/d - 1/ (d+L)]

Using λ = Q/L

E = k Q/L [ 1/d - 1/ (d+L)]

let's use a bit of arithmetic to simplify the expression

[ 1/d - 1/ (d+L)] = L /[d(d+L)]

The final result is

E = k Q / [d(d+L)]