2 added 53 characters in body edited Sep 29 '17 at 9:42 AccidentalFourierTransform 26k1473131 Each element of charge dq$$\mathrm dq$$ on each plate exerts a force on all other elements of charge dq'$$\mathrm dq'$$ on both plates. But dq$$\mathrm dq$$ does not exert a force on itself. Summing the forces from all charges on one plate on any given charge element on that plate gives zero net force. Therefore, the only force on a given charge element is the force from the charge elements on the opposite plate. The same approach applies when calculating the force on a charge element dq$$\mathrm dq$$ at the surface of a spherical shell of charge, due to all other charge elements making up the shell. Setting up the integral explicitly and taking the limit as the contributions from the shell include all elements except the one excluded, we find a field at the boundary of magnitude $$\sigma/(2\epsilon_0)$$, or half what one would obtain if using the field just outside the shell, due to the entire shell. The key point in both problems is the same: Namely, each charge element dq$$\mathrm dq$$ acts like a point charge, and point charges do not exert forces on themselves. The resulting factor of 1/2$$1/2$$ is counterintuitivecounter-intuitive, but it is correct. A different way to obtain the same results is to consider the change in the assembly energy of the distributions under virtual displacements of the excluded charge elements. See Jackson, 3rd edition, pp. 42-43, for a terse but accurate explanation. Each element of charge dq on each plate exerts a force on all other elements of charge dq' on both plates. But dq does not exert a force on itself. Summing the forces from all charges on one plate on any given charge element on that plate gives zero net force. Therefore, the only force on a given charge element is the force from the charge elements on the opposite plate. The same approach applies when calculating the force on a charge element dq at the surface of a spherical shell of charge, due to all other charge elements making up the shell. Setting up the integral explicitly and taking the limit as the contributions from the shell include all elements except the one excluded, we find a field at the boundary of magnitude $$\sigma/(2\epsilon_0)$$, or half what one would obtain if using the field just outside the shell, due to the entire shell. The key point in both problems is the same: Namely, each charge element dq acts like a point charge, and point charges do not exert forces on themselves. The resulting factor of 1/2 is counterintuitive, but it is correct. A different way to obtain the same results is to consider the change in the assembly energy of the distributions under virtual displacements of the excluded charge elements. See Jackson, 3rd edition, pp. 42-43, for a terse but accurate explanation. Each element of charge $$\mathrm dq$$ on each plate exerts a force on all other elements of charge $$\mathrm dq'$$ on both plates. But $$\mathrm dq$$ does not exert a force on itself. Summing the forces from all charges on one plate on any given charge element on that plate gives zero net force. Therefore, the only force on a given charge element is the force from the charge elements on the opposite plate. The same approach applies when calculating the force on a charge element $$\mathrm dq$$ at the surface of a spherical shell of charge, due to all other charge elements making up the shell. Setting up the integral explicitly and taking the limit as the contributions from the shell include all elements except the one excluded, we find a field at the boundary of magnitude $$\sigma/(2\epsilon_0)$$, or half what one would obtain if using the field just outside the shell, due to the entire shell. The key point in both problems is the same: Namely, each charge element $$\mathrm dq$$ acts like a point charge, and point charges do not exert forces on themselves. The resulting factor of $$1/2$$ is counter-intuitive, but it is correct. A different way to obtain the same results is to consider the change in the assembly energy of the distributions under virtual displacements of the excluded charge elements. See Jackson, 3rd edition, pp. 42-43, for a terse but accurate explanation. 1 answered Sep 29 '17 at 9:00 Jon 111 Each element of charge dq on each plate exerts a force on all other elements of charge dq' on both plates. But dq does not exert a force on itself. Summing the forces from all charges on one plate on any given charge element on that plate gives zero net force. Therefore, the only force on a given charge element is the force from the charge elements on the opposite plate. The same approach applies when calculating the force on a charge element dq at the surface of a spherical shell of charge, due to all other charge elements making up the shell. Setting up the integral explicitly and taking the limit as the contributions from the shell include all elements except the one excluded, we find a field at the boundary of magnitude $$\sigma/(2\epsilon_0)$$, or half what one would obtain if using the field just outside the shell, due to the entire shell. The key point in both problems is the same: Namely, each charge element dq acts like a point charge, and point charges do not exert forces on themselves. The resulting factor of 1/2 is counterintuitive, but it is correct. A different way to obtain the same results is to consider the change in the assembly energy of the distributions under virtual displacements of the excluded charge elements. See Jackson, 3rd edition, pp. 42-43, for a terse but accurate explanation.