Two uniformly charged spherical conductors A | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) When two conductors are connected by a conducting wire,charge flows until their potentials become equal,i.e.,$V_A = V_B$.Since the potential of a

Vedclass · Accepted Answer

$2:1$

Vedclass · Answer

(B) When two conductors are connected by a conducting wire,charge flows until their potentials become equal,i.e.,$V_A = V_B$.Since the potential of a spherical conductor is $V = \frac{KQ}{R}$,we have $\frac{KQ_A}{R_A} = \frac{KQ_B}{R_B}$.This implies $\frac{Q_A}{Q_B} = \frac{R_A}{R_B} = \frac{5 \ mm}{10 \ mm} = \frac{1}{2}$.The electric field at the surface of a sphere is given by $E = \frac{KQ}{R^2}$.Therefore,the ratio of the electric fields is $\frac{E_A}{E_B} = \frac{KQ_A / R_A^2}{KQ_B / R_B^2} = \frac{Q_A}{Q_B} 	imes \left(\frac{R_B}{R_A}ight)^2$.Substituting the ratio of charges: $\frac{E_A}{E_B} = \left(\frac{R_A}{R_B}ight) 	imes \left(\frac{R_B}{R_A}ight)^2 = \frac{R_B}{R_A}$.Given $R_A = 5 \ mm$ and $R_B = 10 \ mm$,we get $\frac{E_A}{E_B} = \frac{10}{5} = \frac{2}{1}$.

Similar Questions

$A$ hollow closed conductor of irregular shape is given some charge. Which of the following statements are correct?

Two identical conductors of copper and aluminium are placed in identical electric fields. The magnitude of induced charge in the aluminium will be

Prove that,if an insulated,uncharged conductor is placed near a charged conductor and no other conductors are present,the uncharged body must be intermediate in potential between that of the charged body and that of infinity.

The charges on two plates of a $10\,\mu F$ capacitor are $5\,\mu C$ and $15\,\mu C$. The potential difference across the capacitor plates is........$V$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module