(a) A charged rod P attracts rod R whereas P repels another charged rod Q. What type of force is developed between Q and R?
(b) Define the capacitance of capacitor. Derive and expression for the capacity at a parallel plate capacitor with a dielectric slab placed in between the plates of capacitor.

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Solution

(a)Q and R will develop attractive forces because R is attracted by P whereas Q is repelled by P.
(b)Capacitance of a capacitor is defined as the ratio of the charge stored on any of the plates of capacitor to the potential between the plates.
$V = E d$
since, $\frac{E_{0}}{E} = K$ (where $K$ is dielectric constant of slab)
$V = \frac{E_{0} d}{K}$
we know, $E_{0} = \frac{q}{ϵ_{0} A}$
$V = \frac{q d}{ϵ_{0} A K}$
$C = \frac{q}{V} = \frac{K ϵ_{0} A}{d}$

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(a) A charged rod P attracts rod R whereas P repels another charged rod Q. What type of force is developed between Q and R?(b) Define the capacitance of capacitor. Derive and expression for the capacity at a parallel plate capacitor with a dielectric slab placed in between the plates of capacitor.

(a) A charged rod P attracts rod R whereas P repels another charged rod Q. What type of force is developed between Q and R?
(b) Define the capacitance of capacitor. Derive and expression for the capacity at a parallel plate capacitor with a dielectric slab placed in between the plates of capacitor.