In the given figure electric field at center $O$ due to section $AB$ of uniformly charged ring is $\overrightarrow E$. What will be electric field at $O$ due to section $ACB$ ?

818-747

  • A

    $\overrightarrow E$

  • B

    $ -\overrightarrow E$

  • C

    $\frac{{\overrightarrow E }}{2}$

  • D

    $ -\frac{{\overrightarrow E }}{2}$

Similar Questions

Two point charges $a$ and $b$, whose magnitudes are same are positioned at a certain  distance from each other with a at origin. Graph is drawn between electric field strength at  points between $a$ and $b$ and distance $x$ from a $E$ is taken positive if it is along the line joining from to be. From the graph, it can be decided that 

A charged cork of mass $m$ suspended by a light string is placed in uniform electric filed of strength $E= $$(\hat i + \hat j)$ $\times$ $10^5$ $NC^{-1}$ as shown in the fig. If in equilibrium position tension in the string is $\frac{{2mg}}{{(1 + \sqrt 3 )}}$ then angle $‘\alpha ’ $ with the vertical is

A ring of charge with radius $0.5\, m$ having a $0.02\, m$ gap, carries a charge of $+1\, C$. The field at the centre is

Two point charges $A$ and $B$ of magnitude $+8 \times 10^{-6}\,C$ and $-8 \times 10^{-6}\,C$ respectively are placed at a distance $d$ apart. The electric field at the middle point $O$ between the charges is $6.4 \times 10^{4}\,NC ^{-1}$. The distance ' $d$ ' between the point charges $A$ and $B$ is..............$m$

  • [JEE MAIN 2022]

Two uniform spherical charge regions $S_1$ and $S_2$ having positive and negative charges overlap each other as shown in the figure. Point $O_1$ and $O_2$ are their centres and points $A, B, C$ and $D$ are on the line joining centres $O_1$ and $O_2$. Electric field from $C$ to $D$