Question

The force exerted by an electric charge at the origin on a charged particle at a point (x, y, z) with position vector r = x, y, z is F(r) = Kr/|r|3 where K is a constant. Find the work done as the particle moves along a straight line from (3, 0, 0) to (3, 2, 5).

Answer 1

$W = K[ {\frac{1}{3} - \frac{1}{\sqrt{38} } ]$

Explanation:

The parametric representation of a line segment joining the points (a,b,c) and (l,m,n) is

r(t) = (1-t) . (a,b,c) + t . (l, m, n)  where t ∈ |0, 1|

So, the parametric representation of a line segment joining the points (3,0,0) and (3,2,5) is

r(t) = (1 - t) . (3,0,0) + t . (3,2,5)  where t ∈ |0, 1|

r(t) = (3(1 - t), 0, 0) + (3t, 2t, 5t)  where t ∈ |0, 1|

r(t) = (3, 2t, 5t)

Given:

$F(x, y, z) = \frac{Kr}{|r|^3} \\\\F(x, y, z) = \frac{K}{(x^2 + y^2 + z^2)^3^/^2} (x, y, z)\\\\F(r(t)) = \frac{K}{(3^2 + (2t)^2 + (5t)^2)^3^/^2} (3, 2t, 5t)\\\\F(r(t)) = \frac{K}{(9 + 29t^2)^3^/^2} (3, 2t, 5t)$

dr = (0, 2, 5) dt

$Work = \int\limits^1_0 {F} \, dr$

$W = \int\limits^1_0 {\frac{K}{(9 + 29t^2)^3^/^2} } (3, 2t, 5t) . (0, 2, 5)\, dt\\ \\ = \int\limits^1_0 {\frac{K ( 4t + 25t)}{(9 + 29t^2)^3^/^2} } \, dt$

$W = \frac{1}{2}\int\limits^1_0 {\frac{K(29t)}{(9 + 29t^2)^3^/^2} } \, dt$

Substitute = 9 + 29t² = u, 92tdt = du

Limit changes from 0→1 to 9 → 38

$W = \frac{K}{2} \int\limits^3_9 {\frac{du}{u^3^/^2} } \,$

On solving this, we get:

$W = K[ {\frac{1}{3} - \frac{1}{\sqrt{38} } ]$

Therefore, work done is $W = K[ {\frac{1}{3} - \frac{1}{\sqrt{38} } ]$