The potential at the surface of a conducting sphere is given by:
\( V = \frac{Kq}{r} \).
Since the two spheres are connected, their potentials will be the same:
\[ \frac{Kq_1}{a} = \frac{Kq_2}{b}. \]
Simplifying:
\[ \frac{q_1}{q_2} = \frac{a}{b}. \]
Final Answer: \( \frac{a}{b} \).
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]
An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :
