Q. A parallel-plate capacitor of capacitance $40\,\mu$F is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant $K=2$. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are
A) 4 mC and 0.2 J
B) 8 mC and 2.0 J
C) 2 mC and 0.4 J
D) 2 mC and 0.2 J
Explanation
[Solution by Dr. Abubacker] The battery stays connected, so $V$ is constant. $\Delta Q=(K-1)CV=40\times10^{-6}\times100=4$ mC and $\Delta U=\frac12(K-1)CV^2=\frac12\times40\times10^{-6}\times10^4=0.2$ J. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.