Question

A dipole is centered at the origin, with its axis along the axis, so that at locations on the axis, the electric field due to the dipole is given by E = (0, 1/4 pi epsilon_0 2qs/gamma^3, 0) v/m The charges making up the dipole are q_1 = +4 nC and q_2 = nC, and the dipole separation is s = 8 mm. What is the potential difference along a path starting at location p_1 = (0, 0.04, 0) m and ending at location P_2 = (0, 0.05, 0) m?

Answer 1

Step-by-step explanation:

Given that:

`q_1 = + 4nC \\ \\ q_2 = -4nC \\ \\ q = 4nC`

`P_1 = (0,0.04,0) m\\\\ P_2 = (0,0.05,0) m`

As
`q_1`
`\text{is nearer to }`
`P_1 \& P_2` then
`q_2,q_1`
`\text{is positive, Thus}`,
`E \limits^(\to)`
`\text{ will be along the positive X-axis.}`

Recall that:

`E ^(\to) = (-dV)/(dr)\implies E_y = (-dV)/(dy) \\ \\ = \int^(P_2)_(P_1)dV = -\int \limits ^((0.05))_((0.04))E_y \ dy \\ \\ \implies (V_(P_2)}-V_((P_1))) = (-2q_5)/(4 \pi \varepsilon _o) \int \limits ^((0.05))_((0.04)) (dy)/(y^3) \\ \\ \implies (V_(P_2)}-V_((P_1))) = (-2q_5)/(4 \pi \varepsilon _o) \Big[ (-1)/(2y^2) \Big]^(0.05)_(0.04) \\ \\ \implies (V_(P_2)}-V_((P_1))) = (q_5)/(4 \pi \varepsilon _o) \Big[(1)/(y^2)\Big] ^(0.05)_(0.04)`

`\\ \\ \implies (V_(P_2)}-V_(P_1)) = ((4* 10^(-9))(8* 10^(-3)))/(4 \pi (8.854 * 10^(-12))) \Big [(1)/((0.05)^2)-(1)/((0.04)^2) \Big ]V \\ \\ (V_(P_2)}-V_(P_1)) = \mathbf{-64.71 \ volts}`