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(A) Let $V_A$ be the potential of the outer sphere (radius $R$,charge $Q$) and $V_B$ be the potential of the inner sphere (radius $r$,charge $q$).The

Vedclass · Accepted Answer

$\frac{1}{4\pi \epsilon_0} \left( \frac{q}{r} - \frac{q}{R} \right)$

Vedclass · Answer

(A) Let $V_A$ be the potential of the outer sphere (radius $R$,charge $Q$) and $V_B$ be the potential of the inner sphere (radius $r$,charge $q$).The potential at the surface of the outer sphere $(A)$ is due to its own charge $Q$ and the charge $q$ on the inner sphere:$V_A = \frac{1}{4\pi \epsilon_0} \frac{Q}{R} + \frac{1}{4\pi \epsilon_0} \frac{q}{R} = \frac{1}{4\pi \epsilon_0} \left( \frac{Q+q}{R} \right)$The potential at the surface of the inner sphere $(B)$ is due to its own charge $q$ and the potential due to the outer sphere (which is constant inside it):$V_B = \frac{1}{4\pi \epsilon_0} \frac{q}{r} + \frac{1}{4\pi \epsilon_0} \frac{Q}{R} = \frac{1}{4\pi \epsilon_0} \left( \frac{q}{r} + \frac{Q}{R} \right)$The potential difference between the spheres is $V_B - V_A$:$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left( \frac{q}{r} + \frac{Q}{R} \right) - \frac{1}{4\pi \epsilon_0} \left( \frac{Q+q}{R} \right)$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left( \frac{q}{r} + \frac{Q}{R} - \frac{Q}{R} - \frac{q}{R} \right)$$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left( \frac{q}{r} - \frac{q}{R} \right)$

Section $A$	Section $B$
$(i)$ Nitrosomonas	$(a)$ $\text{Nitrite} \rightarrow \text{Nitrate}$
$(ii)$ Agrobacterium	$(b)$ $\text{Nitrogen fixing bacteria}$
$(iii)$ Rhizobium	$(c)$ $\text{NO}_3^- \rightarrow \text{N}_2$
$(iv)$ Nitrobacter	$(d)$ $\text{Ammonia} \rightarrow \text{Nitrite}$
	$(e)$ $\text{Denitrifying bacteria}$

$A$ small conducting sphere of radius $r$ is lying concentrically inside a bigger hollow conducting sphere of radius $R$. The bigger and smaller spheres are charged with $Q$ and $q$ $(Q > q)$ respectively and are insulated from each other. The potential difference between the spheres will be

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