Calculating Self Capacitance of an Isolated Sphere

[ehild]

I posted it above in one of my post. :-)

You mean you wrote up Cself in terms of R1, R2, R3 long ago, you just let me work in vain? Good by, Pranav...

 

[Saitama]

You mean you wrote up Cself in terms of R1, R2, R3 long ago, you just let me work in vain? Good by, Pranav...

Did I say something wrong?

$$V(R_2)=kq_1 \left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$
$R_1=10 cm, R_2=30 cm, R_3=40 cm$
Using the expression posted above for $q_1$
$$q_1=Q\frac{R_1(R_3-R_2)}{R_2(R_3-R_1)}$$
$$q_1=Q\frac{10 \times 10}{30 \times 30}$$
$$q_1=\frac{Q}{9}$$

Using this in $V(R_2)$
$$V=k \frac{Q}{9} \cdot \frac{R_2-R_1}{R_1 R_2}$$
$$V=9 \times 10^9 \times \frac{Q}{9} \times \frac{20 \times 100}{10 \times 30}$$
Solving.
$$V(R_2)=\frac{20 \times 10^9 Q}{3}$$

$$C_{self}=\frac{Q}{V(R_2)}$$
$$C_{self}=\frac{3Q}{20 \times 10^9 Q}$$
Solving this, I get $1.5 \times 10^{-10}$
Is this correct?

 

[ehild]

yes.

It is equal the the parallel resultant of two spherical shell capacitors.

ehild

 

[Saitama]

yes.
It is equal the the parallel resultant of two spherical shell capacitors.

Thanks a lot ehild!

But why the self capacitance is equal to the parallel combination here?

 

[ehild]

I can not read the mind of your teacher, why he/she called the thing "self" capacitance. Anyway, you had something carrying charge and having some potential with respect to the ground, which was proportional to the charge. So it had capacitance C=Q/V.
ehild

 

[Saitama]

I can not read the mind of your teacher, why he/she called the thing "self" capacitance. Anyway, you had something carrying charge and having some potential with respect to the ground, which was proportional to the charge. So it had capacitance C=Q/V.

This question was not by my teacher. :P

Thanks once again! :)