Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surface of spheres A and b?

Answers (1)

Electric Field Intensity -

\vec{E}=\frac{\vec{F}}{q_{0}}=\frac{kQ}{r^{2}}

- wherein

 

 

Capacitance of Conductor -

Q\propto V

Q=CV

- wherein

C - Capacity or capacitance of conductor 

V - Potential.

 

 When the spherical conductors are connected by a conducting wire, charge is redistributed and the spheres attain a common potentia  V.

\therefore\: \: \: Intensity \: E_{A}= \frac{1}{4\pi \varepsilon _{0}}\frac{Q_{A}}{R_{A}^{2}}

or\: \: \: \: E_{A}= \frac{1\times C_{A}V}{4\pi \varepsilon _{0}R_{A}^{2}}= \frac{\left ( 4\pi \varepsilon _{0}R_{A} \right )V}{4\pi \varepsilon _{0}R_{A}^{2}}= \frac{V}{R_{A}}

Similarly  E_{B}= \frac{V}{R_{B}}

\therefore \frac{E_{A}}{E_{B}}= \frac{R_{B}}{R_{A}}= \frac{2}{1}

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Knockout JEE Main January 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions