Question

1. A spherical balloon is inflated to a radius of R. It is given a charge of Q so that it has a uniform surface charge density of σ. At some point, P, fairly far from the balloon the electric field strength due to the balloon is measured to be E.

(a) The balloon is electrically isolated so that no charge can flow onto or off of it. While it is isolated it is inflated to a radius of 2R. What is the new surface charge density of the balloon? What is the new total charge on the balloon?

(b) After the balloon is inflated as described above what is the electric field strength due to the balloon at point P?

(c) The balloon is deflated so that it again has a radius of R and it is again charged with charge Q so that it has a uniform charge density of σ. Now it is inflated while in contact with a charged object. This causes the charge density on the balloon to remain constant while it is being inflated (in practice it would be very hard to arrange for this to happen). The balloon is inflated to a radius of 2R. The other charged object is then removed. What is the new charge on the balloon?

(d) After the balloon is inflated as described in c), what is the electric field strength due to the balloon at point P?

Answer 1

a)

Total new charge = Q_new = Total old charge = Q

before the balloon is inflated :

= Q/A = Q/(4R²)

after the balloon is inflated , r = radius = 2 R

' = Q_new/A_new = Q/(4r²) = Q/(4(2R)²) = (1/4) Q/(4R²) = (0.25)

b)

Since the total charge on the balloon remain same after being inflated, the electric field at point P also remain same .hence electric field = E

c)

when radius is "R" , charge on balloon is given as

Q = (4R²)

when radius is "2R" , charge on balloon is given as

Q' = (4(2R)²) = 4 (4R²) = 4 Q

so new charge on the balloon is 4 times

d)

since the charge on the balloon has becomes 4 times , and we know that electric field is directly propotional to the charge. hence the electric field at point P also becomes 4 times

so electric field = 4 E