If closed integral of E.dS is zero over a surface, then...?

[Dev]

Homework Statement

If closed integral of E.dS is zero over a surface, then...
Option a) the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
Option b) all charges must necessarily be outside the surface.

Relevant Equations

Gauss Law states that closed integral of E.dS is equal to enclosed charge upon permittivity of free space.

I think only Option a should be correct because net electric flux can be zero even if charges are inside the surface. For example, if two charges of +2 coulomb and -2 coulomb are inside the surface, then also net electric flux over the surface is zero!

But my textbook, as well as various online sources say that both options a and b are correct. Why is b correct?

 

[PhDeezNutz]

I agree with you.

What level class are you in? Is it Freshman University level?

That’s probably why they gave you a naive answer……if it’s a first introduction things need to be presented in a way that is easier than the whole truth.

Still I think that’s ridiculous of them to do.

 

[Dev]

I agree with you.

What level class are you in? Is it Freshman University level?

That’s probably why they gave you a naive answer.

I am in Class 12, India. The question is from NCERT Exemplar Class 12.

 

[Orodruin]

As stated here b is incorrect. However, this may be an issue of your interpretation of the problem or an issue of the problem itself. Please reproduce the problem statement exactly as posed (or provide a reference).

 

[Dev]

As stated here b is incorrect. However, this may be an issue of your interpretation of the problem or an issue of the problem itself. Please reproduce the problem statement exactly as posed (or provide a reference).

 

[Dev]

Both options C and D are given to be correct in the answer key.

 

[Orodruin]

(d) is definitely false.

 

[Orodruin]

To expand on that:

The total charge inside the surface must be zero. That does not mean that there are no charges inside the surface. You could, for example have one point particle of charge $2q$ and two of charge $-q$. The net charge inside the surface, and therefore the flux integral, is then zero.

 

[Dev]

Thanks! I don't understand why do they make such errors and other online sources just copy that blindly!

 

[kuruman]

(d) is definitely false.

Also the answer key is incorrect and it's the word "necessarily" that makes it so. If this were a multiple answer problem (provide all answers that could be correct) and not a multiple choice question (provide the only correct answer), then the answer key would be correct.

Thanks! I don't understand why do they make such errors and other online sources just copy that blindly!

I would say they make such errors because of carelessness or indifference. A less charitable word would be ignorance.

 

[Dev]

But this question is indeed a multiple -answer problem.

 

[kuruman]

But this question is indeed a multiple -answer problem.

Then (d) would also be correct if it read
(d) all charges are outside the surface
without "must" and "necessarily."

 

[Orodruin]

If this were a multiple answer problem (provide all answers that could be correct) and not a multiple choice question (provide the only correct answer), then the answer key would be correct.

I don’t think multiple choicd vs multiple answer has anything to do with this. You can very well have a multiple answer question asking you to select all correct options with only one correct option. (Or all options correct, or zero)

 

[PeroK]

Then (d) would also be correct if it read
(d) all charges are outside the surface
without "must" and "necessarily."

"If ... then" must be read as a logical implication, where the statement in the question implies the answer. You're reading it the other way round, where the answer is correct if it implies the statement in the question.

That said, the question is poorly phrased by not making it clear that all potential answers are on the same footing, and must be implied by the statement in the question to be counted as correct.

 

[kuruman]

"If ... then" must be read as a logical implication, where the statement in the question implies the answer. You're reading it the other way round, where the answer is correct if it implies the statement in the question.

That said, the question is poorly phrased by not making it clear that all potential answers are on the same footing, and must be implied by the statement in the question to be counted as correct.

You are right. I attempted to salvage the question without the full benefit of my morning stimulant concoction.