A) 2
B) 1
C) 4
D) 8
Correct Answer: C
Solution :
\[{{C}_{1}}={{\varepsilon }_{0}}\frac{A}{{{d}_{1}}}\] and \[{{C}_{2}}=K{{\varepsilon }_{0}}\frac{A}{{{d}_{2}}}\] \ \[\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{1}{K}\times \frac{{{d}_{2}}}{{{d}_{1}}}\] \[=\frac{C}{2C}=\frac{1}{K}\times \frac{2d}{d}\] Þ K = 4You need to login to perform this action.
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