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Rom4ik [11]
3 years ago
8

We want to design a cylindrical vacuum capacitor, with a given radius a for the outer cylindrical shell, that will be able to st

ore the greatest amount of electrical energy per unit length, subject to the constraint that the electric field strength at the surface of the inner cylinder may not exceed E0. What radius b should be chosen for the inner cylindrical conductor, and how much energy can be stored per unit length
Physics
1 answer:
anastassius [24]3 years ago
8 0

Solution :

a). Using Gauss's law :

  $E=\frac{Q}{4 \pi \epsilon_0r^2}$  ,    $b    .........(1)

Let $E=E_0,\ r=b$ in equation (1)

Therefore, $Q=4 \pi \epsilon_0b^2E_0$  .............(2)

$V_b-V_a = \int^a_b \vec E. d\vec l$

             $=\int^a_b E \ dx$

            $=\frac{Q}{4 \pi \epsilon_0} \int^a_b \frac{1}{x^2} \ dx$

            $=\frac{Q(a-b)}{4 \pi \epsilon_0 a b}$  ....................(3)

Therefore, $U=\frac{1}{2}Q \Delta V$

                     $=\frac{1}{2}(4 \pi \epsilon_0 b^2 E_0)\left(\frac{Q(a-b)}{4\pi \epsilon_0 a b}\right)$

                     $=\frac{4 \pi \epsilon_0}{2a} \ E^2_0 b^3(a-b)$  .............(4)

Now differentiating the equation (4) w.r.t. 'b', we get

$b=\frac{3}{4}a$  

Thus the radius for the inner cylinder conductor is $b=\frac{3}{4}a$

b). For the energy storage, substitute the radius in (4), we get

$U = 4 \pi\epsilon_0 \frac{27a^3E^2_0}{512}$

This is the amount of energy stored in the conductor.

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