A) increase
B) decrease
C) remain same
D) be zero
Correct Answer: A
Solution :
\[{{F}_{air}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] \[{{F}_{glass}}=\frac{1}{4\pi {{\varepsilon }_{0}}K}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}=\frac{{{F}_{air}}}{K}\] \[\therefore \] \[{{F}_{glass}}>{{F}_{air}}\]
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