A) 3
B) 4
C) 5
D) 6
Correct Answer: C
Solution :
[c] Before introducing a slab capacitance of plates \[{{C}_{1}}=\frac{{{\varepsilon }_{0}}A}{3}\] If a slab of dielectric constant K is introduced between plates then \[C=\frac{K{{\varepsilon }_{0}}A}{d}\text{ then }{{C}_{1}}'=\frac{{{\varepsilon }_{0}}A}{2.4}\] \[{{C}_{1}}\] and \[{{C}_{1}}'\]are in series hence, \[\frac{{{\varepsilon }_{0}}A}{3}=\frac{k\frac{{{\varepsilon }_{0}}A}{3}.\frac{{{\varepsilon }_{0}}A}{2.4}}{k\frac{{{\varepsilon }_{0}}A}{3}+\frac{{{\varepsilon }_{0}}A}{2.4}}\]\[3k=2.4k+3\text{ }0.6k=3\] Hence, the dielectric constant of slap in given by, \[k=\frac{30}{6}=5\]You need to login to perform this action.
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