The spatial distribution of the electric fiel | vedclass

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Question

(A) Electric field lines originate from a positive charge and terminate at a negative charge. In the figure,the lines are originating from $A$ and ter

Vedclass · Accepted Answer

$A$ is $+ve$ and $B$ is $-ve$ and $|A| > |B|$

Vedclass · Answer

(A) Electric field lines originate from a positive charge and terminate at a negative charge. In the figure,the lines are originating from $A$ and terminating at $B$,so $A$ is positive $(+ve)$ and $B$ is negative $(-ve)$.Furthermore,the number of electric field lines originating from or terminating at a charge is proportional to the magnitude of the charge. By counting the lines,we see that more lines originate from $A$ than terminate at $B$. Therefore,the magnitude of charge $A$ is greater than the magnitude of charge $B$,i.e.,$|A| > |B|$.

The spatial distribution of the electric field due to charges $A$ and $B$ is shown in the figure. Which one of the following statements is correct?

Similar Questions

$A$ charge $Q$ is enclosed by a Gaussian surface of radius $R$. If the radius is doubled,then the outward electric flux will

$A$ uniformly charged conducting sphere of $2.4\; m$ diameter has a surface charge density of $80.0\; \mu C/m^2$.
$(a)$ Find the charge on the sphere.
$(b)$ What is the total electric flux leaving the surface of the sphere?

The linear charge density of a wire is $8.85 \ \mu C/m$. The radius and height of the cylinder are $3 \ m$ and $4 \ m$ respectively. Find the electric flux passing through the cylinder.

The figure shows electric field lines due to a charge configuration. From this,we conclude that:

$A$ uniformly charged conducting sphere of diameter $3.5 \ cm$ has a surface charge density of $20 \ \mu C \ m^{-2}$. The total electric flux leaving the surface of the sphere is nearly:
[permittivity of free space,$\epsilon_0 = 8.85 \times 10^{-12} \ SI \ unit$]

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