I understand everything except one thing... Why is the electric field radial ?

to my understanding because the charge is unformly spread across the surface. and when we did the integration for the electric field of a uniformly charged surface. at any point the field was perpendicularto the surface. if we take very small pieces of the surface they will be planar too and thus why we have radial electric field. i kinda had the same doubt and this is the best ive been able to come up with,somebody correct me please if im wrong.

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Course: Class 12 Physics (India) > Unit 1

Lesson 11: Application of Gauss law

Electric field due to spherical shell of charge

Google Classroom

Electric field due to a charged spherical shell

Part 1- Electric field outside a charged spherical shell

Let's calculate the electric field at point

P

, at a distance

r

from the center of a spherical shell of radius

R

, carrying a uniformly distributed charge

Q

Khan Academy video wrapper

Field due to spherical shell of charge See video transcript

Why is the electric field radial, outside the sphere?

[Task] - Come up with an argument to prove that the electric field must be radial & only depend on the distance from the center, $r$ ‍.

STEP 1 - Choosing a Gaussian surface

Now that we know what the electric field looks like everywhere, choose a Gaussian surface that would make calculating the electric flux, easy.

What closed surface would you choose here?

Now that we have chosen a closed surface, we are ready to apply Gauss law & calculate the electric field.

STEP 2 - Simplifying the left-hand side (L.H.S.) of Gauss's law

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SphericalCharge(Part4)See video transcript

Gauss law L.H.S

Let's simplify the equation. Can you try simplifying the L.H.S,

\oint \vec{E} . \vec{d A}

\oint \vec{E} . \vec{d A}

Note 1: Input your answer in terms of $E$ ‍ & $r$ ‍.
Note 2: On desktop, type 'pi' for $π$ ‍ and press 'ctrl+^' to input exponents.

STEP 3 - Simplifying the right-hand side (R.H.S.) of Gauss's law.

Gauss law R.H.S

Now let's try simplifying the R.H.S,

\frac{q_{i n s}}{ϵ_{0}}

q_{i n s} / ϵ_{0} =

ϵ_{0}

STEP 4 - Equating L.H.S & R.H.S

Now we can equate both sides of Gauss's law & calculate the electric field strength at point

P

Equating LHS & RHS

E

Note: On desktop, input $e$ ‍ for $ϵ_{0}$ ‍ and 'pi' for $π$ ‍

Putting it all together

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SphericalChargeEField(Part5)See video transcript

Part 2 - Electric field inside the charged spherical shell

Khan Academy video wrapper

Field inside (Part 6)See video transcript

STEP 2 - Simplifying the L.H.S. of Gauss's law

Gauss law L.H.S

Let's simplify the L.H.S,

\oint \vec{E} . \vec{d A}

\oint \vec{E} . \vec{d A}

Note 1: Input your answer in terms of $E$ ‍ & $r$ ‍.
Note 2: On desktop, type 'pi' for $π$ ‍ and press 'ctrl+^' to input exponents.

STEP 3 - Simplifying the R.H.S. of Gauss's law

Gauss law R.H.S

Now let's try simplifying the R.H.S,

\frac{q_{i n s}}{ϵ_{0}}

q_{i n s} / ϵ_{0} =

ϵ_{0}

STEP 4 - Equating L.H.S & R.H.S

Now we can equate both sides of Gauss's law & calculate the electric field strength inside the shell.

Equating LHS & RHS

E

Note: On desktop, input $e$ ‍ for $ϵ_{0}$ ‍ and 'pi' for $π$ ‍

Let's summarise

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Part7FinalSummaryFieldduetoSphericalShellSee video transcript

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aniketprasad123
Posted 3 years ago. Direct link to aniketprasad123's post “I understand everything e...”
I understand everything except one thing...
Why is the electric field radial ?
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(9 votes)
Answer
- dyoffis
  Posted 2 years ago. Direct link to dyoffis's post “to my understanding becau...”
  to my understanding because the charge is unformly spread across the surface. and when we did the integration for the electric field of a uniformly charged surface. at any point the field was perpendicularto the surface.
  if we take very small pieces of the surface they will be planar too and thus why we have radial electric field.
  i kinda had the same doubt and this is the best ive been able to come up with,somebody correct me please if im wrong.
  Button navigates to signup page
  (7 votes)
Ahmed E.
Posted 3 years ago. Direct link to Ahmed E.'s post “But we learned that E at ...”
But we learned that E at a point is = k q/r^2 r-hat. How do we now have a different definition for E?
Edit: just found that k = 1 / 4 pi epsilon. which is how it's written in the vids
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(4 votes)
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