#### Thank you for registering.

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

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
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

# Charge q is distributed uniformly on an arc of radius R subtending right angle at its centre.Another charge -q is placed at the centre of arc.Find dipole moment of system??

10 years ago

consider a small element on arc of length dl substending d@ at the center ...

let angle made with base is @ ...this is placed in XY plane with base in x axis ...

charge of dl element = dq

dq = pRd@             (p is charge dencity)

let us assume that central charg  divided into n parts of dq magnitude then every dq element of arc

& central charge will constitute a dipole ...

dP (dipole moment) = q(2a) = dq R

dP = dqR2 [ cos@(i) + sin@ (j) ] lim from 0 to pi/2

dP = pR2[ cos@d@(i) + sin@d@(j) ]      lim 0 to pi/2

P(net dipole moment) = pR2 [ i + j ]

now , p = q/(piR/2) = total charge/length of arc so

Pnet = 2qR/pi . [ i + j]   with direction

5 months ago
if the net dipole moment is to be calculated with a charge q distributed uniformly on an arc of radius R subtending right angle at its centre. Another charge -q placed at the centre of the arc.

Then the net dipole moment to be calculated with distributed dipole is

Pnet = 2qR/pi sin(pi/2)