Question

A straight uniform rod of charge Q = 40.0 nC lies on the x-axis from x = 0 to x = 4.0 m. Determine the net electric field at x = 6.0 m on the x axis.

Answer 1

Answer:

Electric Field of a Line Segment Find the electric field a distance z above the midpoint of a straight line segment of length L that carries a uniform line charge density .

Strategy Since this is a continuous charge distribution, we conceptually break the wire segment into differential pieces of length dl, each of which carries a differential amount of charge . Then, we calculate the differential field created by two symmetrically placed pieces of the wire, using the symmetry of the setup to simplify the calculation ((Figure)). Finally, we integrate this differential field expression over the length of the wire (half of it, actually, as we explain below) to obtain the complete electric field expression.

A uniformly charged segment of wire. The electric field at point P can be found by applying the superposition principle to symmetrically placed charge elements and integrating.

Explanation:

point charge:

line charge:

Answer 2

A straight uniform rod of charge Q = 40.0 nC lies on the x-axis from x = 0 to x = 4.0 m.

We have to find the net electric field at x = 6 m , on the x-axis.

What is Electric field ?

Electric field is a region around an electrically charged particle or object in which it feels electric force. it is defined as the electric force per unit charge. i.e., E = $\frac{F}{q}=\frac{kqQ}{r^2q}=\frac{kQ}{r^2}$

here, Q = charge of particle = 40 nC = 40 × 10⁻⁹ C

length of straight uniform rod, l = 4 m

so the linear charge density of rod, λ = Q/l = 10⁻⁸ C/m

we know, we can apply above mentioned formula only when the charge is point charge. but here charge is linear straight uniform rod. so we have to find electric field using integration.

cut an element of thickness dx,

charge of this element would be, dQ = λdx

now electric field due to this charge at point x = 6 m , dE = kdQ/x²

⇒ $\int\limits^E_0dE=\int\limits^{x_2}_{x_1}{\frac{k\lambda}{x^2}},dx$

⇒ E = $k\lambda\left[-\frac{1}{x}\right]^{x_2}_{x_1}$

here, x₂ = maximum distance of the element from the point of observation = 6 m

x₁ = minimum distance of the element from the point of observation = 6 - 4 = 2 m

⇒ E = 9 × 10⁹ × 10⁻⁸ [1/2 - 1/6]

= 90 × 1/3 = 30 N/C

Therefore the net electric field due to straight rod is 30 N/C.