A charge of 4.4 10-6 C is located in a uniform electric field of intensity 3.9 105 N/C. How much work (in Joules) is required to move this charge 50 cm along a path making an angle of 40

The work required to move this charge is 0.657 J

Explanation:

Given;

magnitude of charge, q = 4.4 x 10⁻⁶ C

Electric field strength, E =  3.9 x 10⁵ N/C

distance moved by the charge, d = 50 cm = 0.5m

angle of the path, θ = 40°

Work done is given as;

W = Fd

W = FdCosθ

where;

F is the force on the charge;

According the coulomb's law;

F = Eq

F = 3.9 x 10⁵ x 4.4 x 10⁻⁶  = 1.716 N

W = FdCosθ

W = 1.716 x 0.5 x Cos40

W = 0.657 J

Therefore, the work required to move this charge is 0.657 J