Question

A conducting wire connects two charged metallic spheres \(A\) and \(B\) of radii \(r_1\) and \(r_2\) respectively. The distance between the spheres is very large compared to their radii. The ratio of electric fields, \((E_A/E_B)\), at the surfaces of spheres \(A\) and \(B\) will be

• A. \(\dfrac{r_1}{r_2}\)
• B. \(\dfrac{r_2}{r_1}\)
• C. \(\dfrac{r_1^2}{r_2^2}\)
• D. \(\dfrac{r_2^2}{r_1^2}\)

Hint:

For conducting spheres connected by a wire: \[ V_A=V_B \] and since \[ E=\frac{V}{R}, \] the electric field at the surface of a sphere is inversely proportional to its radius.

Answer 1

The Correct Option is B

Concept: When two conducting spheres are connected by a conducting wire, charge flows between them until both spheres attain the same electric potential. For an isolated conducting sphere of radius \(R\), \[ V=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R} \] and the electric field at its surface is \[ E=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}. \] Since the potentials of the connected spheres become equal, we can first establish the relation between their charges and then determine the ratio of electric fields.

Step 1: Apply the condition of equal potentials.
Since the spheres are connected, \[ V_A=V_B \] Therefore, \[ \frac{1}{4\pi\varepsilon_0}\frac{Q_A}{r_1} = \frac{1}{4\pi\varepsilon_0}\frac{Q_B}{r_2} \] or, \[ \frac{Q_A}{r_1}=\frac{Q_B}{r_2} \] Hence, \[ \frac{Q_A}{Q_B} = \frac{r_1}{r_2}. \]

Step 2: Write expressions for electric fields at the surfaces.
\[ E_A=\frac{1}{4\pi\varepsilon_0}\frac{Q_A}{r_1^2} \] and \[ E_B=\frac{1}{4\pi\varepsilon_0}\frac{Q_B}{r_2^2} \] Therefore, \[ \frac{E_A}{E_B} = \frac{Q_A}{Q_B} \times \frac{r_2^2}{r_1^2} \] Substituting \[ \frac{Q_A}{Q_B} = \frac{r_1}{r_2}, \] we get \[ \frac{E_A}{E_B} = \frac{r_1}{r_2} \times \frac{r_2^2}{r_1^2} = \frac{r_2}{r_1}. \] Thus, \[ \boxed{\frac{E_A}{E_B}=\frac{r_2}{r_1}} \]