Question

A steady current of 30 micro ampere flows between the two electrodes of an X-ray tube for 10 s. If the potential difference between the electrodes be 40 kV, what is the work done in transferring the charge from one electrode to the other?​

Answer 1

Explanation:

W=neV=> work done in transferring charge of 'n' number of electrons

V= 40×10^3 V

I = Q/t

Q = It = 30×10^-6 ×10 = 3×10^-4

Since Q=ne

W= ne*V

W= 3×10^(-4)×40×10^(3)

W = 12 J

Answer 2

Explanation:

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