Question

A uniform electric field is directed along the +x axis and a point charge q= +6.0 nC is moving in the electric field. First, th charge travels 40cm along the +x axis and then 30 cm along the -y axis. Find the magnitude of the electric field if the total work done by the field in moving the charge is W= 3.6 x 10^-6 J.

Answer 1

Here are two cases

i) when particle is travel along x- axis we know in this case displacement of charge is in +x direction which is in the direction of electric force/field

Workdone 1 is Force× displacement = charge × Electric field × displacement

6 × 10^-9 × E × 40 × 10 ^-2 = WORKDONE 1.

ii) when particle is travel along y axis then angle between force and displacement is 90 degree hence work done is zero in 2nd case.

So W-1 + W-2 = 3.6 × 10 ^-6

24× 10^-10 × E = 3.6 × 10 ^-6

E = 0.15× 10^+4 V/M