Force on one half due to the other

Electricity and Magnetism Level 4

Two half-spheres with fixed surface charge densities $\sigma_1$ and $\sigma_2$ and radius $R$ are brought into contact, as shown to the right.

Find the force of interaction between the two half-spheres.

\frac{\sigma_1 \sigma_2}{8\epsilon_0} \pi R^2

\frac{\sigma_1 \sigma_2}{6\epsilon_0} \pi R^2

\frac{\sigma_1 \sigma_2}{4\epsilon_0} \pi R^2

\frac{\sigma_1 \sigma_2}{2\epsilon_0} \pi R^2

2 solutions

Navin Murarka
Nov 4, 2017

Consider a small patch of area $dA_1$ and $dA_2$ on the two hemispheres.The force of interaction between two hemispheres $F=\int \int \frac{\sigma_1 dA_1 \sigma_2 dA_2}{4\pi \epsilon_0 |r_{12}|^2 } \hat{r_{12}}$ which can be written as $F=\frac{\sigma_1 \sigma_2}{4\pi \epsilon_0} k$ where $k=\int \int \frac{dA_1dA_2}{|r_{12}|^2} \hat{r_{12}}$ is a purely geometric quantity.Here the double integral runs over the surface area of the sphere.

In order to evaluate the geometrical quantity use $\sigma$ as the uniform charge density of the entire sphere.So $\frac{\sigma^2}{2\epsilon_0} \pi R^2 = \frac{\sigma^2}{4\pi \epsilon_0} k$ Obtain k and subsitute to get the required force of interaction

@Navin Murarka ,

Nice solution but can't i take two elemental hollow spheres from both halves and then calculate the elemental force on one due to other by concentrating them at their respective centres?

In nutshell, can i avoid double integrations?

Priyanshu Mishra - 3 years, 6 months ago

Numerical integration tells me that your answer is correct, but I don't follow your solution. I understand the first few lines, but what does it mean "use $\sigma$ as the uniform charge density of the entire sphere..." and so on?

Bradley Treece - 3 years, 6 months ago

Es correcta la solución, pero hay que usar conceptos de energía para hallar la fuerza entre las dos semiesférica cargadas uniformemente con sigma

Moisés Sanchez Arteaga - 3 years, 6 months ago

Can you explain how do you get the final equality involving k and R ? Does it come from the electrostatic energy of the distribution, Gauss theorem or what ?

Vincent Cauchois - 3 years, 6 months ago

It comes from this.consider a small pill box inside the sphere.we know inside electric field is zero.so the electric field just outside the sphere is $EA=\frac{\sigma A\}{\epsilon_0}$ so $E=\frac{\sigma}{\epsilon_0}$ This electric field is due to contribution from local charge inside the pill box and the rest of the sphere.inside the contribution from both of them are zero.So $E_{rest}=\frac{\sigma}{2\epsilon_0}$ thus the charge in the pill box experience a pressure $\frac{\sigma^2 A}{2\epsilon_0}$

Navin Murarka - 3 years, 6 months ago

Ding Yuchen
Dec 10, 2017

The interaction between any two small pieces from each hemisphere is always proportional to $\sigma_1 \times \sigma_2$ , so the force between the two hemispheres is also proportional to it. Knowing this, we can

set $\sigma_1 = \sigma_2 \equiv \sigma$ ,
evaluate the interaction force (which will be proportional to $\sigma^2$ ),
and then restore the general result by replacing $\sigma^2$ with $\sigma_1 \sigma_2$ .

For the case where $\sigma_1 = \sigma_2 \equiv \sigma$ , the system becomes a uniformly charged sphere, with charge $Q$ and potential energy $\mathcal{E}$ $Q = 4 \pi R^2 \sigma \\ \mathcal{E} = \frac{Q^2}{8 \pi \epsilon_0 R}$ The key is to evaluate the tension $f$ on the charged sphere, which can be obtained using the principal of virtual work .

Consider an infinitesimal change of radius $R \rightarrow R + dR$ , the change of the potential energy will be $d\mathcal{E} = - \frac{Q^2 dR}{8 \pi \epsilon_0 R^2}$ which should match the work done by the tension force on the sphere $dW = f \times d A = f dR \frac{d}{dR} ( 4 \pi R^2 ) = 8 \pi f R dR \equiv - d \mathcal{E}$ where $A = 4 \pi R^2$ is the area of the sphere.

The the interaction force between the two hemisphere (when $\sigma_1 = \sigma_2 \equiv \sigma$ ) is $F = 2\pi R \times f = \frac{Q^2} {32 \pi \epsilon_0 R^2} = \frac{\pi R^2 \sigma^2}{2 \epsilon_0}$

Replacing $\sigma^2$ with $\sigma_1 \sigma_2$ gives the final result.

@Ding Yuchen ,

Wow this is simply brilliant.

But i have some queries about the very first line of your solution. How it comes?

Priyanshu Mishra - 3 years, 6 months ago

i too have the same doubt

subrahmanyam subrahmanyam - 3 years, 4 months ago

Mathematically it's derived in Navin Murarka's answer. The $\sigma_1\sigma_2$ can be taken out of the integral. By intuition, if $\sigma_1$ increases by 10%, then: every piece on the left has 10% more charge; every piece on the right feels 10% more force from each piece on the left; (since the force vectors are added linearly:) the magnitude of the force one piece on the right get from all the pieces on the left increases by 10%, with direction unchanged; the magnitude of the total force the right part get from the the left increases by 10%, with direction unchanged. The force between both part is proportional to $\sigma_1$ , and the same applies to $\sigma_2$ .

Ding Yuchen - 3 years, 2 months ago

Force on one half due to the other

2 solutions

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