Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

A non-conducting disc of positive surface charge density 's' is placed horizontally on the ground. A particle of mass 'm' is dropped along the axis of the disc from height 'H' with 0 initial velocity. The particle has q/m=pi*g/sk . Find the height H such that the particle just reaches the disc.

A non-conducting disc of positive surface charge density 's' is placed horizontally on the ground. A particle of mass 'm' is dropped along the axis of the disc from height 'H' with 0 initial velocity. The particle has q/m=pi*g/sk . Find the height H such that the particle just reaches the disc.

Grade:12

1 Answers

Saurabh Kumar
askIITians Faculty 2411 Points
6 years ago
The electric potential of the disk at points along its axis needs to be determined first. We can then equate the potential energy (gravitational plus electrical) at height H to that at the disk's center, because "just reaching the disk" means that initial and final kinetic energies are zero.

Adding (integrating) contributions to the potential from infinitesimally thin rings making up the disk gives for a point on the axes at height z:

V(z) = k ∫ s 2π r dr /sqrt(r^2 + z^2)
= ks 2π ( sqrt(R^2 + z^2) - z)

So the potential energy for the particle with charge q at height H is

U = q V(H)
= 2π q k s ( sqrt(R^2 + H^2) - H )

Therefore equalling the potential energies at initial and final positions gives

m g H + 2π q k s ( sqrt(R^2 + H^2) - H )= 2π q k s R

This we have to solve for H:
2π q k s ( sqrt(R^2 + H^2) - H ) = 2π q k s R - m g H

sqrt(R^2 + H^2) - H = R - mgH/(2π q k s )

It is given that q k s = πmg, So

sqrt(R^2 + H^2) = R + H(1-1/(2π^2) )

Squaring both sides gives

R^2 + H^2 = R^2 + 2 R H (1-1/(2π^2)) + H^2 (1-1/2π^2)^2

H^2 (1 - (1-1/2π^2)^2 )= 2 R H (1-1/(2π^2))


H = 0 or H = 2 R (1-1/(2π^2))/(1 - (1-1/2π^2)^2 )

So

H = 2R (1-1/(2π^2)) / (1/π^2 - 1/(4π^4))
= 2R (4π^4 -2π^2)/(4π^2-1)
= 4π^2 (2π^2-1)/(4π^2-1) R

which is about 19.2 R

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free