A point charge of magnitude 2μC is moved through a distance of 0.02 m against uniform field of intensity 25 Vm^-1 Calculate the work done on the charge

daddybarbie

7 years ago

workdone =Qv : w =Qv :

but E =v/d : v = 25 x 0.02= 0.5v :

w = 2 x 10^—6 × 0.5 = 1 ×10^—6J

Deeto

3 years ago

Work Done= ½QV
where Q = 2 microFarad = 2x10^-6F
V = Er = 25 x 0.02 = 0.5 (r is distance, E is electric field intensity)
Therefore, work done = ½x2x10^-6x0.05
Work done = 5 x 10^-7 J
Option C ✓ is supppsed to be the correct answer. Please make corrections

Sikirublaze1995

11 years ago

Dont take anything has been simple, please solve for us and make it simple and let the solvings be corresct

BrodaChoco

1 year ago

Intensity E = Force/ charge
Force = E × charge
Work done = distance × force
Work done = distance × E × charge
Work done = 0.02m × 25J/m × 2E-6c
Work done = 1E-6Jc or VC

ChideraJohnAgu

2 months ago

the right answer is C
½QV = work done
V =0.02×25=0.5
½×2×10^–6 ×0.5= 5×10^-7