#### 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

# a soap bubble of radius 3 cm is charged with 9 nC . the excess pressure inside the bubble , if surface tension of soap solution is 3× 10-3 N/m.ans is .36 N/m2

Arun
25763 Points
one year ago
dear stdent
excess pressre due to surface tension =4S/r
here S=3*10−3 r= 3cm excess pressre = 4SR=4*3*10−3/3*10−2 = 0.4now excess ressure due to electric field = σ22ε=q22A2εq22(4πr2)ε here ε=8.85*10−12 here q= 9nm=9*10−9C r=3cm =3*10−2putting all values we get pressure due to electric field= 0.035N/m2its is outward to the spherical bubble so net pressure= 0.4− 0.035 =0.36N/m2
Vikas TU
14149 Points
one year ago
dear stdent
excess pressre due  to  surface tension =4S/r
here  S=3*10^−3 r= 3cm excess pressre = 4S/R
=4*3*10^−3/3*10^−2
= 0.4now excess ressure due to electric field
= σ^2/2ε=q^2/2A^2ε
q^2/2(4πr^2)ε
here ε=8.85*10^−12
here q= 9nm=9*10^−9C
r=3cm =3*10−2putting all values we get pressure due to electric field= 0.035N/m2
its is outward to the spherical bubble so net pressure= 0.4− 0.035 =0.36N/m2