The capacitors A and B have identical geometry. A material with a dielectric constant 3 is present b

Question

The capacitors A and B have identical geometry. A material with a dielectric constant 3 is present between the plates of B . The potential difference across $A$ and $B$ are respectively\

Professor PC Thomas Classes · Answer

A and B are connected in series. Same charge flows through them\ $\mathrm{C}_{\mathrm{A}}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}} ; \mathrm{C}_{\mathrm{B}}=\frac{3 \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}=3 \mathrm{C}_{\mathrm{A}}$\ $	herefore$ Resultant capacity $C=\frac{C_{A} 	imes C_{B}}{C_{A}+C_{B}}$\ or $C=\frac{C_{A} 	imes 3 C_{A}}{C_{A}+3 C_{A}}=\frac{3 C_{A}}{4}$\ $	herefore$ Net charge $=$ CV\ $\mathrm{q}=\frac{3}{4} \mathrm{C}_{\mathrm{A}} 	imes 10=7.5 \mathrm{C}_{\mathrm{A}}$\ $	herefore \mathrm{V}_{\mathrm{A}}=\frac{\mathrm{q}}{\mathrm{C}_{\mathrm{A}}}=7.5 \mathrm{C}_{\mathrm{A}}$\ $	herefore \mathrm{V}_{\mathrm{A}}=\frac{\mathrm{q}}{\mathrm{C}_{\mathrm{A}}}=7.5 \mathrm{C}_{\mathrm{A}}$\ $\mathrm{V}_{\mathrm{B}}=\frac{\mathrm{q}}{\mathrm{C}_{\mathrm{B}}}=\frac{\mathrm{q}}{3 \mathrm{C}_{\mathrm{A}}}=2.5$ volt\