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 $ACB$ ?

The given diagram shows two semi infinite line of charges having equal (in magnitude) linear charge density but with opposite sign. The electric field at any point on $x$ axis for $(x > 0)$ is along the unit vector

A positively charged thin metal ring of radius $R$ is fixed in the $xy - $ plane with its centre at the $O$. A negatively charged particle $P$ is released from rest at the point $(0,\,0,\,{z_0})$, where ${z_0} > 0$. Then the motion of $P$ is

Electric field strength due to a point charge of $5\,\mu C$ at a distance of $80\, cm$ from the charge is

The electric field due to a charge at a distance of $3\, m$ from it is $500\, N/coulomb$. The magnitude of the charge is.......$\mu C$ $\left[ {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}\,\frac{{N - {m^2}}}{{coulom{b^2}}}} \right]$

Infinite charges of magnitude $q$ each are lying at $x =1,\, 2,\, 4,\, 8...$ meter on $X$-axis. The value of intensity of electric field at point $x = 0$ due to these charges will be