In a vacuum diode, electrons are “boiled” off a hot cathode, at potential zero, and accelerated across a gap to the anode, which is held at positive potential $V_{0}$ . The cloud of moving electrons within the gap (called space charge) quickly builds up to the point where it reduces the field at the surface of the cathode to zero. From then on a steady current I flows between the plates. Suppose the plates are large relative to the separation $({A>>d}^{2})$ , so that edge effects can be neglected.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

The value of potential V at a distance x from cathode is given by $V (x) = V_{0} \frac{x}{d}$

The value of potential V at a distance x from cathode is given by $V (x) = V_{0} {(\frac{x}{d})}^{4 / 3}$

The value of steady current I between the plates will be proportional to $V_{0}^{4 / 3}$

The value of steady current I between the plates will be proportional to $V_{0}^{3 / 2}$

answer is B, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

I = uAv = constant
Here v is speed.
Using poisson’s equation,
$\frac{d^{2} v}{d x^{2}} = \frac{- u}{E_{0}} = - \frac{I}{E_{0} A v} v = \sqrt{\frac{2 q V}{m}} \frac{d^{2} v}{d x^{2}} = \frac{- Im}{E_{0} A \sqrt{2 q V}} = S v^{- \frac{1}{2}}$
Multiplying $v'$ both sides.
$v' \frac{d v'}{d x} = - S v^{- \frac{1}{2}} v'$
Solving with proper limits
$v = v_{0} {(\frac{x}{d})}^{4 / 3}$