How Does Electrostatic Force Affect Charges Within a Uniformly Charged Sphere?

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The discussion focuses on calculating the electrostatic force within a uniformly charged sphere, specifically on the northern hemisphere. The total force exerted by the electrostatic field can be derived using Coulomb's Law and integrating over the volume of the hemisphere. For a sphere of radius R containing a total charge Q, the force is expressed as F = (2/5) * (9 * 10^9) * (Q^2 / (4/3 * pi * R^3)) * pi^2 * R^5. Evaluations for specific cases, such as using R=1m and 10kg of electrons, as well as the repulsive force between two such spheres separated by 3000km, are also discussed. The calculations highlight the significant difference between electrostatic and gravitational forces acting on the charged masses.
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Consider a solid sphere of radius R containing a total charge Q which is uniformly distributed throughout the volume with a volume charge density rho.

a) find the total force exerted by the electrostatic field on the charge in the 'northern hemisphere'. express your answer in terms of the total charge Q and the radius R of the sphere

b) evaluate your answer to part a) using R=1m containing 10kg of electrons

c) find the repulsive force between two sperical volumes of radius R=1m, each containing 10kg of electrons held fixed and separated by a distance of 3000km

d) by what factor does the magnitude of the repulsive electro. force, found in part c) exceed the gravitational attractive force of the two masses

e) if the forces holding the two spherical volumes of electrons, fixed at a distance of 3000km, were to vanish, find the initial acceleration that would be experienced by the two 10kg masses. express your answer of units of g-forces of 9.81m per second squared.
 
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The most straightforward way to sove your problems is to use Maxwell's stress tensor
to find the appropriate forces. I'm sure I've seen the first problem done in textbooks, but the specifics escape me.
Regards,
Reilly Atkinson
 


a) To find the total force exerted by the electrostatic field on the charge in the northern hemisphere, we can break it down into smaller infinitesimal charges and integrate over the volume. The force exerted by an infinitesimal charge dq located at a distance r from the center of the sphere is given by Coulomb's Law: dF = k * (Q * dq) / r^2, where k is the Coulomb's constant. Since the charge is uniformly distributed, we can express dq in terms of the volume charge density rho as dq = rho * dV, where dV is the volume element. The volume element in spherical coordinates is given by dV = r^2 * sin(theta) * dr * dtheta * dphi. Integrating over the northern hemisphere (theta from 0 to pi/2 and phi from 0 to 2*pi), we get:

F = ∫∫∫ dF = ∫∫∫ k * (Q * rho * r^2 * sin(theta) * dr * dtheta * dphi) / r^2
= k * Q * rho * ∫∫∫ r * sin(theta) * dr * dtheta * dphi
= k * Q * rho * ∫∫ r^2 * sin(theta) * dtheta * dphi
= k * Q * rho * ∫ 0 to 2*pi ∫ 0 to pi/2 ∫ 0 to R r^2 * sin(theta) * r^2 * sin(theta) * dr * dtheta * dphi
= k * Q * rho * (2*pi) * ∫ 0 to pi/2 (sin(theta))^2 * ∫ 0 to R r^4 * dr
= k * Q * rho * (2*pi) * (pi/2) * (R^5 / 5)
= (2/5) * k * Q * rho * pi^2 * R^5

Substituting k = 9 * 10^9 N*m^2/C^2, Q = total charge, and rho = Q / (4/3 * pi * R^3), we get:

F = (2/5) * (9 * 10^9) * (Q / (4/3 * pi * R^3)) * Q * pi^2 *
 
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