Ncert Exercises & Solutions

Question 2.20.

Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.

Let a be the radius of sphere A, Q_A be the charge on the sphere, and C_A be the capacitance of the sphere.

Let b be the radius of sphere B, Q_B be the charge on the sphere, and C_B be the capacitance of the sphere.

Since the two spheres are connected with a wire, their potential (V) will become equal.

Let E_A be the electric field of sphere A and E_a be the electric field of sphere B. Therefore, their ratio, And

\begin{aligned} \frac{E_{A}}{E_{B}} = \frac{Q_{d}}{4 {πϵ}_{0} \times a_{2}} \times \frac{b^{2} \times 4 {πϵ}_{0}}{Q_{B}} \\ \frac{E_{A}}{E_{B}} = \frac{Q_{A}}{Q_{g}} \times \frac{b^{2}}{a^{2}} . . . (1) \\ However, \frac{Q_{i}}{Q_{s}} = \frac{C_{d} V}{C_{B} V} \\ And, \frac{C_{A}}{C_{B}} = \frac{a}{b} \\ ∴ \frac{Q_{d}}{Q_{n}} = \frac{a}{b} . . . (2) \end{aligned}

putting the value of (2) in (1), we obtain

∴ \frac{E_{A}}{E_{B}} = \frac{a}{b} \frac{b^{2}}{a^{2}} = \frac{b}{a}

Therefore, the ratio of electric fields at the surface is b/a.

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