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): \(2:1\)
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
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).