Consider Gauss-s law - E - ds - q -0q 2Then- for the situation shown in figure at the Gaussian surfaceA- E- due to q 2 would be zeroB- E- due to both q 1 and q 2 would be non-zeroC- - due to both q 1 and q 2 would be non-zeroD- - due to q 2 would be zero

The correct options are B E due to both q_1 and nbsp;q_2 nbsp; would be non zero D ϕ due to q_2 would be zeroLet us first draw the electric field lines comi ...

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Byju's Answer

Standard XII

Physics

Electric Field Due to a Line of Charge Not on Its Axis

Consider Gaus...

Question

Consider Gauss's law $\oint \to E . \to d s = \frac{q}{\in_{0}}$

Then, for the situation shown in figure at the Gaussian surface

\to E

due to

q_{2}

would be zero

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\to E

due to both

q_{1}

and

q_{2}

would be non-zero

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ϕ

due to both

q_{1}

and

q_{2}

would be non-zero

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ϕ

due to

q_{2}

would be zero

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Solution

The correct options are
B $\to E$ due to both $q_{1}$ and $q_{2}$ would be non-zero
D $ϕ$ due to $q_{2}$ would be zero
Let us first draw the electric field lines coming out of charges $q_{1}$ and $q_{2}$

As the electric field lines coming out of charge $q_{2}$ are entering the gaussian surface so the electric field due to $q_{2}$ would be non-zero.

In case of charge $q_{1}$ also the electric field due charge $q_{1}$ are crossing the gaussian surface so the electric field due to $q_{2}$ would also be non-zero.

From the above figure we can say that due to charge $q_{2}$
The number of field lines entering the gaussian surafce = The number of field lines leaving the gaussian surface
$∴ ϕ$ due to $q_{2} = 0$
The charge $q_{1}$ is placed inside the gaussian surface which means that the eletric field lines are only leaving the gaussian surface and not entering inside it.
$⟹ ϕ$ due charge $q_{1}$ will be non-zero.

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Similar questions

Q. The electric force on a charge

q_{1}

due to

q_{2}

(4^i - 3^j) N

. The unit vector in the direction of electirc force on

q_{2}

due to

q_{1}

will be:

Q. The electric force on a charge

q_{1}

due to

q_{2}

(4 ˆ i - 3 ˆ j)

N. What is the force on

q_{2}

due to

q_{1}

q_{1}, q_{2}, q_{3} a n d q_{4}

are point charges located at points as shown in the figure and S is spherical Gaussian surface of radius R. Which of the following is true according to the Gauss’s law.

Q. Change

Q_{1}

and

Q_{2}

lie inside and outside respectively of a closed surface

S

. Let

E

be the field at any point on

S

ϕ

be the flux of

E

(electric field) over

S

	List-I		List-II
(P)	If $Q_{1} = 0$ and $Q_{2} \neq 0$ then	(1)	Both $E$ and $ϕ$ will change
(Q)	If $Q_{1} \neq 0$ and $Q_{2} = 0$ then	(2)	$E \neq 0$ and $ϕ \neq 0$
(R)	If $Q_{2}$ changes	(3)	net electric field will change but $ϕ$ will not change
(S)	If $Q_{1}$ changes	(4)	$E \neq 0$ but $ϕ = 0$

Q. Consider the situation shown in the figure. What are the signs of q₁ and q₂? If the lines are drawn in proportion to the charges, what is the ratio q₁/q₂?

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Consider Gauss's law ∮→E.→ds=q∈0 Then, for the situation shown in figure at the Gaussian surface

Consider Gauss's law $\oint \to E . \to d s = \frac{q}{\in_{0}}$

Then, for the situation shown in figure at the Gaussian surface