Two uniformly charged spherical conductors, A and B of radii 5 mm and 10 mm are separated by a distance of 2 cm. If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitude of the electric fields at surface of the spheres A and B will be
- 1:2
- 2:1
- 1:1
- 1:4
The Correct Option is B
Solution and Explanation
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): \(2:1\)
Top Questions on Electric Field
- An electron is released in the field generated by a non-conductivity sheet of uniform surface charge density $ \sigma $. The rate of change of de-Broglie wavelength associated with the electron varies inversely as the $ n $th power of the distance travelled. Find the value of $ n $.
- JEE Main - 2025
- Physics
- Electric Field
- A potential at a point A is 3 V and that at another point B is 5 V. What is the work done in carrying a charge of 5 mC from B to A?
- KCET - 2025
- Physics
- Electric Field
Two large plane parallel conducting plates are kept 10 cm apart as shown in figure. The potential difference between them is $ V $. The potential difference between the points A and B (shown in the figure) is:
- JEE Main - 2025
- Physics
- Electric Field
- Work done to move a unit charge from one point to the other in an electric circuit is called
- AP POLYCET - 2025
- Physics
- Electric Field
- Electric field lines around a positive charge are directed:
- KEAM - 2025
- Physics
- Electric Field
Questions Asked in JEE Main exam
- Consider the following reactions: \[ \text{NH}_2 \text{O} \xrightarrow{\text{NaNO}_2, \text{HCl}, 0-5^\circ C} [A] \xrightarrow{\text{HCl dil.}, \text{NaOH}} [B] \] - (i) The molecular formula of [A] is C₁₄H₁₄N₂O₂
- (ii) The molecular formula of [C] is C₁₆H₁₈N₂O₂ The total number of sp³ hybridized carbon atoms in the major product [C] formed is -----.- JEE Main - 2025
- Organic Reactions
If the four distinct points $ (4, 6) $, $ (-1, 5) $, $ (0, 0) $ and $ (k, 3k) $ lie on a circle of radius $ r $, then $ 10k + r^2 $ is equal to
- JEE Main - 2025
- Coordinate Geometry
- Consider a complex reaction taking place in three steps with rate constants \(k_1\), \(k_2\), and \(k_3\) respectively. The overall rate constant \(k\) is given by the expression \( k = \sqrt{\frac{k_1 k_3}{k_2}} \). If the activation energies of the three steps are 60, 30, and 10 kJ mol\(^{-1}\) respectively, then the overall energy of activation in kJ mol\(^{-1}\) is ________________(Nearest integer).
- JEE Main - 2025
- Thermodynamics
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is:
- JEE Main - 2025
- Number Systems
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
- JEE Main - 2025
- Organic Chemistry
Concepts Used:
Electric Field
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).