A spherical balloon contains a charge +Q uniformly distributed over its surface. When it has a diameter D, the electric field at its surface

Question

A spherical 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?

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3 years 2021-08-17T04:07:20+00:00 1 Answers 42 views 0

E = 1/9  E₀

Explanation:

In this exercise we are told that the electric field is Eo when the diameter of the balloon is D, the expression

we are asked to shorten the electric field when the diameter is 3D with the same eclectic charge

For this we can use the gauss law to find the field in the new diameter, for this we create a Gaussian surface in the form of a sphere

Ф = ∫ E. dA = /ε₀

In this case the lines of the electric field and the radii of the sphere are parallel, therefore the scalar product is reduced to the algebraic product and the charge inside the sphere is the initial charge Q

A = 4π r²

E 4π r² = Q /ε₀

E = 1 /4πε₀     Q / r²

the value of the indicated distance is 3 times the initial diamete

r  = 3 D / 2

we substitute

E = 1/4 πε₀ Q (2/ 3D)²

for the initial conditions

E₀ = 1 / 4πε₀ Q  (2/D)²

subtitled in the equation above

E = 1/9  E₀