The gravitational field in a region is given by $E = (\to 2 i + \to 3 j) N {k g}^{- 1}$ . Show that no work is done by the gravitational field when a particle is moved on the line $3 y + 2 x = 5$ .

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Solution

Electric field acts in direction when slope is $m_{1} = \frac{3}{2}$
But slope of the line $3 y + 2 x = 5$ is $m_{2} = \frac{- 2}{3}$
Force is along the direction of field , $(m_{1} \times m_{2} = (\frac{3}{2} \times \frac{- 2}{3}) = - 1$
Angle between electric filed direction and displacement direction is,
$tan θ = \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}$ = $\infty$
$θ = 90$
Since $θ = 90$ ,no work is done by the field.

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The gravitational field in a region is given by E=(→2i + →3j)N kg−1. Show that no work is done by the gravitational field when a particle is moved on the line 3y+2x=5.

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The gravitational field in a region is given by $E = (\to 2 i + \to 3 j) N {k g}^{- 1}$ . Show that no work is done by the gravitational field when a particle is moved on the line $3 y + 2 x = 5$ .