Question

A conducting sphere of radius R is cut into two equal halves which are held pressed together by a stiff spring inside the sphere. (a) Find the change in tension in the spring if the sphere is given a charge Q. (b) Find the change in tension in the spring corresponding to the maximum charge that can be placed on the sphere if dielectric breakdown strength of air surrounding the sphere is $E_0$.

Answer 1

(a) $\frac{Q^2}{32\pi R^2 \epsilon_0}$ (b) $\frac{\pi \epsilon_0 R^2 E_0^2}{2}$

Answer 2

The problem asks us to calculate the change in tension in a spring that holds two hemispherical conducting shells together when the sphere is charged.

Part (a): Find the change in tension in the spring if the sphere is given a charge Q.

1. Charge Distribution and Electric Field: When a conducting sphere of radius R is given a charge Q, the charge distributes uniformly over its outer surface.
   The surface charge density is $\sigma = \frac{Q}{4\pi R^2}$.
   The electric field just outside the surface of the conductor is $E = \frac{\sigma}{\epsilon_0}$.

2. Electrostatic Pressure: Due to the repulsion between like charges, there is an outward electrostatic pressure on the surface of the conductor. This pressure is given by:
   $P = \frac{1}{2}\epsilon_0 E^2$
   Substitute $E = \frac{\sigma}{\epsilon_0}$:
   $P = \frac{1}{2}\epsilon_0 \left(\frac{\sigma}{\epsilon_0}\right)^2 = \frac{\sigma^2}{2\epsilon_0}$
   Now substitute $\sigma = \frac{Q}{4\pi R^2}$:
   $P = \frac{1}{2\epsilon_0} \left(\frac{Q}{4\pi R^2}\right)^2 = \frac{Q^2}{32\pi^2 R^4 \epsilon_0}$

3. Force of Separation: This pressure acts radially outward. To find the total force tending to separate the two hemispheres, we need to find the component of this pressure acting perpendicular to the cutting plane (the plane that divides the sphere into two halves). This can be calculated by multiplying the pressure by the projected area of one hemisphere onto the cutting plane.
   The projected area of a hemisphere onto its base is a circle of radius R, so its area is $A_{proj} = \pi R^2$.
   The total electrostatic force ($F_{sep}$) pushing the two halves apart is:
   $F_{sep} = P \times A_{proj} = \left(\frac{Q^2}{32\pi^2 R^4 \epsilon_0}\right) (\pi R^2)$
   $F_{sep} = \frac{Q^2}{32\pi R^2 \epsilon_0}$

4. Change in Tension: The spring initially holds the two halves together. When the sphere is charged, the electrostatic force $F_{sep}$ pushes the halves apart. To counteract this force and keep the halves together, the spring must exert an additional force (or its compression must decrease by this amount). The change in tension in the spring is equal to this electrostatic force.
   $\Delta T = F_{sep} = \frac{Q^2}{32\pi R^2 \epsilon_0}$

Part (b): Find the change in tension in the spring corresponding to the maximum charge that can be placed on the sphere if dielectric breakdown strength of air surrounding the sphere is $E_0$.

1. Maximum Charge ($Q_{max}$): The maximum charge that can be placed on the sphere is limited by the dielectric breakdown strength of the surrounding air, $E_0$. This means the electric field at the surface of the sphere cannot exceed $E_0$.
   The electric field at the surface of a charged sphere is $E = \frac{Q}{4\pi\epsilon_0 R^2}$.
   For maximum charge, $E_{max} = E_0$:
   $E_0 = \frac{Q_{max}}{4\pi\epsilon_0 R^2}$
   Therefore, $Q_{max} = 4\pi\epsilon_0 R^2 E_0$

2. Change in Tension with Maximum Charge: Substitute $Q_{max}$ into the expression for $\Delta T$ derived in part (a):
   $\Delta T_{max} = \frac{Q_{max}^2}{32\pi R^2 \epsilon_0}$
   $\Delta T_{max} = \frac{(4\pi\epsilon_0 R^2 E_0)^2}{32\pi R^2 \epsilon_0}$
   $\Delta T_{max} = \frac{16\pi^2\epsilon_0^2 R^4 E_0^2}{32\pi R^2 \epsilon_0}$
   $\Delta T_{max} = \frac{\pi \epsilon_0 R^2 E_0^2}{2}$

The final answer is $\boxed{\text{(a) }\frac{Q^2}{32\pi R^2 \epsilon_0} \text{ (b) } \frac{\pi \epsilon_0 R^2 E_0^2}{2}}$.

Explanation of the solution: (a) The charge Q distributes uniformly on the sphere's surface, creating an outward electrostatic pressure $P = \frac{Q^2}{32\pi^2 R^4 \epsilon_0}$. The force tending to separate the two hemispheres is this pressure multiplied by the projected area of a hemisphere onto the cutting plane, which is $\pi R^2$. This yields a separation force of $\frac{Q^2}{32\pi R^2 \epsilon_0}$, which is the change in spring tension. (b) The maximum charge $Q_{max}$ that can be placed on the sphere is determined by the breakdown strength $E_0$ of air, such that $E_0 = \frac{Q_{max}}{4\pi\epsilon_0 R^2}$. Substituting $Q_{max} = 4\pi\epsilon_0 R^2 E_0$ into the expression for tension change from part (a) gives $\frac{\pi \epsilon_0 R^2 E_0^2}{2}$.