Spheres after connecting with a conducting wire :
\(σ_1R_1 = σ_2R_2\)
Then, \(E = \frac{σ }{ ∈_0}\)
Hence, the ratio of the magnitude of the electric fields at surface of the spheres A and B will be :
\(⇒ \frac{E_1}{E_2}\)
= \(\frac{σ_1}{σ_2}\)
= \(\frac{R_2}{R_1}\)
= \(\frac{2}{1}\)
Hence, the correct option is (B): \(2:1\)
A metallic ring is uniformly charged as shown in the figure. AC and BD are two mutually perpendicular diameters. Electric field due to arc AB to O is ‘E’ magnitude. What would be the magnitude of electric field at ‘O’ due to arc ABC?
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).