a-where-a-l-which-integral-correctly-gives-the-x-component-of-the-electric-field

Question

A uniformly charged rod lies along the xx-axis from x=0x = 0 to x=Lx = L with linear charge density λ\lambda. At a point on the xx-axis located at x=ax = a, where  L"\u003ea\u003eLa \u003e L, which integral correctly gives the xx-component of the electric field?

Accepted Answer

Ex=kλ∫0Ldx(a−x)2E_{x}=k\lambda\$$int_0$$^{L}\frac{\mathrm{d}x}{\left(a-x\$$right)^2$$}

Answer

Ex=kλ∫0Ldx(a+x)2E_{x}=k\lambda\$$int_0$$^{L}\frac{\mathrm{d}x}{\left(a+x\$$right)^2$$}

Answer

Ex=kλ∫0Lxdx(a−x)2E_{x}=k\lambda\$$int_0$$^{L}\frac{x\mathrm{d}x}{\left(a-x\$$right)^2$$}

Answer

Ex=kλ∫0L(a−x)2dxE_{x}=k\lambda\$$int_0$$^{L}\left(a-x\$$right)^2$$\mathrm{d}x

A uniformly charged rod lies along the xx-axis from x=0x = 0 to x=Lx = L with linear charge density λ\(\lambda\). At a point on the xx-axis located at x=ax = a, where a>La > L, which integral correctly gives the xx-component of the electric field?

A uniformly charged rod lies along the $x$ -axis from $x equals 0$ to $x = L$ with linear charge density $λ$ . At a point on the $x$ -axis located at $x = a$ , where $a > L$ , which integral correctly gives the $x$ -component of the electric field?