In the given circuit,if point C is connected | vedclass

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(C) The circuit consists of two capacitors of $5 \, \mu F$ and $15 \, \mu F$ connected in series between points $A$ and $C$.Given that point $C$ is ea

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$500$

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(C) The circuit consists of two capacitors of $5 \, \mu F$ and $15 \, \mu F$ connected in series between points $A$ and $C$.Given that point $C$ is earthed,its potential $V_C = 0 \, V$. The potential at point $A$ is $V_A = 2000 \, V$.The potential difference across the series combination is $V_{AC} = V_A - V_C = 2000 - 0 = 2000 \, V$.For capacitors in series,the potential difference across a capacitor is inversely proportional to its capacitance $(V \propto 1/C)$.The potential at point $B$ (between the two capacitors) can be found using the potential divider rule for capacitors:$V_B = V_A - V_1 = V_A - \left( \frac{C_2}{C_1 + C_2} ight) V_{AC}$Here,$C_1 = 5 \, \mu F$ and $C_2 = 15 \, \mu F$.$V_B = 2000 - \left( \frac{15}{5 + 15} ight) 	imes 2000$$V_B = 2000 - \left( \frac{15}{20} ight) 	imes 2000$$V_B = 2000 - 1500 = 500 \, V$.

In the given circuit,if point $C$ is connected to the earth and a potential of $+2000 \, V$ is applied to point $A$,what is the potential of point $B$ in volts?

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