Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

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}

Posted by

Satyajeet Kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’
anam-khan

NIT Puducherry Cut-off 2024: Computer Science and Engineering had highest closing rank last year
anam-khan

JEE Mains: Farmer's son from remote Maharashtra village aces exam

Latest Question