Question

A spherical e/3 balloon contains a charge +Q uniformly distributed over its surface. When it has a diameter D, the electric field at its surface has magnitude E. If the balloon is not blown up to thrice this diameter without changing the charge, the electric field at its surface is:________.

A. 4E.
B. 2E.
C. E.
D. E/2.
F. E/4.

Answer 1

Complete Question

A spherical e/3 balloon contains a charge +Q uniformly distributed over its surface. When it has a diameter D, the electric field at its surface has magnitude E. If the balloon is not blown up to thrice this diameter without changing the charge, the electric field at its surface is:________.

A. 4E.

B. 2E.

C. E.

D. E/2.

E E/9

F. E/4.

Answer:

The correct option is E

Step-by-step explanation:

From the question we are told that

The charge on the spherical balloon is
`+Q`

The diameter of the spherical balloon is D

The magnitude of the electric field at the surface is E

Generally the electric field at the surface of the sphere is mathematically represented as

`E = (k Q)/(r ^2)`

where r is the radius which can be represented as

`r = (d)/(2)`

So
`E = (k Q)/([(d)/(2) ] ^2)`

`E = \frac{k *Q * 4} {d^2}`

Now when the diameter become thrice its original values the magnitude of the electric field becomes

`E_n = (k Q * 4)/((3d)^2)`

`E_n = (k Q * 4)/(9 d^2)`

=>
`E_n =(1)/(9) E`