Calculating the Electric Potential Difference of a Helium Balloon

[Jchem]

I'm learning about charge right now and I have no idea how to start this one.. any help is appreciated.


A helium balloon has a charge of q = 5.5 x 10^-8. It rises vertically into the air. d = 600m. from the surface of the Earth to final position A. The electric field that normally exists in the atmosphere near the surface of the Earth has a magnitude E = 150 N/C and is directed downward. What is the difference in electric potential between the two positions?

 

[Doc Al]

What's the definition of electric potential? How does it relate to the field? (Hint: The charge on the balloon is irrelevant.)

 

[Jchem]

The electric field that normally exists in the atmosphere near the surface of the Earth has a magnitude E = 150 N/C and is directed downward.

Is this true for d = 0m and d = 600m ?

if so i think I could solve it using:

V= Ed

 

[Doc Al]

Is this true for d = 0m and d = 600m ?

Yes, assume the field is uniform.

if so i think I could solve it using:

V= Ed

Right!

A more accurate way to write this is:
$$\Delta V = - E \Delta d$$

 

[Dr.Brain]

I note that, even though the electric field is pointing downwards, the positively charged balloon moves upwards, unless and until it decelerates and comes to rest but that a different case.