Along straight metal rod has a radius of 5.00 cm anda charge per unit length of 30.0 nc/m. find the electric field (a)3.00 cm, (b) 10.0 cm, (c) 100cm from the axis of the rod, wheredistances are measured perpendicular to the rod.

given,

radius of rod = 5 cm

charge per unit length (λ) = 30.0 n C/m = 30 × 10⁻⁹ C/m

electric field duel to long straight wire =

a) distance = 3 + 5 = 8 cm

= 18 × 10⁹

E = 6750 N/C

b) distance = 10 + 5 = 15 cm

= 18 × 10⁹

E = 3600 N/C

c) distance = 100 + 5 = 105 cm

= 18 × 10⁹

E =514.28 N/C

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weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards