For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant $K$ is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4} d$, where $'d'$ is the separation between the plates of parallel plate capacitor. The new capacitance $(C')$ in terms of original capacitance $\left( C _{0}\right)$ is given by the following relation

Q: For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant $K$ is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4} d$, where $'d'$ is the separation between the plates of parallel plate capacitor. The new capacitance $(C')$ in terms of original capacitance $\left( C _{0}\right)$ is given by the following relation

c $C _{0}=\frac{\in_{0} A }{ d }$ $C ^{\prime}= C _{1}$ and $C _{2}$ in series. i.e. $\frac{1}{ C ^{\prime}}=\frac{1}{ C _{1}}+\frac{1}{ C _{2}}$ $\frac{1}{ C ^{\prime}}=\frac{(3 d / 4)}{\in_{0} KA }+\frac{ d / 4}{\in_{0} A }$ $\frac{1}{ C ^{\prime}}=\frac{ d }{4 \in_{0} A }\left(\frac{3+ K }{ K }\right)$ $C ^{\prime}=\frac{4 K C _{0}}{(3+ K )}$981-s542

Question

A$C ^{\prime}=\frac{3+ K }{4 K } C _{0}$
B$C ^{\prime}=\frac{4+ K }{3} C _{0}$
C$C ^{\prime}=\frac{4 K }{ K +3} C _{0}$
D$C ^{\prime}=\frac{4}{3+ K } C _{0}$

JEE MAIN 2021, Medium

Download our app
and get started for free

Similar Questions

Download our appand get started for free

Similar Questions

Download our app
and get started for free