Solving for potential using surface charge density of a sphere

In summary, the process of solving for the electric potential due to a sphere involves using the surface charge density, which is defined as charge per unit area on the sphere's surface. By applying Gauss's law, one can determine the electric field both inside and outside the sphere. The potential is then calculated by integrating the electric field with respect to distance, taking into account the symmetry of the sphere. This approach allows for a clear understanding of how charge distribution affects the electric potential in different regions surrounding the sphere.
  • #1
BuggyWungos
13
1
Homework Statement
I'm trying to solve for the potential of a charged copper sphere with only radius and electric field strength known. The field lines are directed into the sphere.
Radius of the sphere: 0.2 m
Electric Field Strength at the surface of the sphere: 3800 N/C
Answer = half of my solution's value.
Relevant Equations
rho = Q/A
Surface area = 4pi(r^2)
Electric Field strength = rho/2(epsilon nought) OR kQ/r^2
Electric potential = kQ/r
surfafce area = 0.502

E = -q/A2(en) = 3800
-q = 3800*(A2(en))
-q = 1.68*10^(-8)
-q = 3.37*10^(-8)

V = kq/r
V = (9.0*10^9)(-3.37*10^(-8))/0.2
V = -1519 V
 
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  • #2
BuggyWungos said:
Electric Field strength = rho/2(epsilon nought)
This is not the correct formula for the electric field at the surface. This is where your mistake of a factor of 2 occurs.
 
  • #3
TSny said:
This is not the correct formula for the electric field at the surface. This is where your mistake of a factor of 2 occurs.
What is the correct electric field strength formula using rho? I understand that E = rho/(epsilon nought) would give the correct answer, but the formula I was given in my textbook was E = 2(pi)k(rho), which would simplify to E = rho/2(epsilon nought). Is the above formula used for another situation?
 
  • #4
BuggyWungos said:
What is the correct electric field strength formula using rho? I understand that E = rho/(epsilon nought) would give the correct answer, but the formula I was given in my textbook was E = 2(pi)k(rho), which would simplify to E = rho/2(epsilon nought). Is the above formula used for another situation?
gives the field of an infinite plane with uniform surface charge density . (The symbol is more often used for a volume charge density rather than a surface charge density.)

The field at a point just outside the surface of a conductor in electrostatic equilibrium is . This can be derived using Gauss’ law.
 
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FAQ: Solving for potential using surface charge density of a sphere

What is surface charge density?

Surface charge density is defined as the amount of electric charge per unit area on a surface. It is typically denoted by the symbol σ (sigma) and is measured in coulombs per square meter (C/m²).

How do you calculate the electric potential due to a charged sphere?

The electric potential V at a distance r from the center of a uniformly charged sphere can be calculated using the formula V = k * Q / r for points outside the sphere, where k is Coulomb's constant (approximately 8.99 x 109 N m²/C²) and Q is the total charge of the sphere. For points inside the sphere, the potential is constant and equal to the potential at the surface.

What is the relationship between surface charge density and total charge?

The total charge Q on a sphere can be determined from the surface charge density σ and the surface area A of the sphere using the formula Q = σ * A. For a sphere, the surface area is given by A = 4πr², so Q = σ * 4πr².

How does the electric potential behave inside and outside the charged sphere?

Outside the charged sphere, the electric potential decreases with distance from the sphere according to the formula V = k * Q / r. Inside the sphere, the electric potential remains constant and is equal to the potential at the surface, as there is no electric field within a uniformly charged sphere.

What assumptions are made when solving for potential using surface charge density?

When solving for potential using surface charge density, it is typically assumed that the sphere is uniformly charged, meaning that the charge is distributed evenly over the surface. It is also assumed that the sphere is isolated, and external electric fields or charges do not influence the potential calculation.

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