capacitor

27 small drops , having charge q and radius r coalesce to form a bigger drop.How many times capacitance will become?


The answer is 3.

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1 solution

Chew-Seong Cheong
Aug 13, 2014

Capacitance C C is given by: C = Q V C=\frac{Q}{V} , where Q Q and V V are the amount of charge and voltage involved.

The voltage V V of a conducting sphere is given by: V = Q 4 π ϵ 0 R V=\frac{Q}{4\pi\epsilon_0R} , where ϵ 0 \epsilon_0 is the permittivity of space and R R is the radius of the sphere.

Therefore, the capacitance C C of a sphere is: C = Q V = 4 π ϵ 0 C=\frac{Q}{V}=4\pi\epsilon_0 .

We note that volume of a sphere V s V_s is proportional to radius r 3 r^3 or V s r 3 V_s \propto r^3 .

Therefore, V s 1 V s 0 = r 3 r 1 3 = 27 1 r 1 3 = 27 r 3 r 1 = 3 r \frac{V_{s1}}{V_{s0}}=\frac{r^3}{r_1^3}=\frac{27}{1} \Rightarrow r_1^3=27r^3 \Rightarrow r_1=3r .

The new capacitance C 1 = 4 π ϵ 0 r 1 = 4 π ϵ 0 ( 3 r ) = 3 C 0 C_1 = 4\pi \epsilon_0r_1= 4\pi \epsilon_0(3r)=3C_0 . Therefore, capacitance becomes 3 \boxed{3} times of before.

There is an error. It should be C = Q V = 4 π ϵ 0 R C = \frac{Q}{V}=4\pi \epsilon_0 R

Chew-Seong Cheong - 6 years, 10 months ago

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