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What is the electric field Ē. at a point outside the cell 3.25 µm from the surface?

What is the electric field Ē. at a point outside the cell 3.25 µm from the surface?

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Please help solve for E4

Red blood cells can be modeled as spheres of 7.03 µm diameter with -2.55 × 10-12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space eo is 8.85 × 10-12 C/(V-m). What is the electric field Éj inside the cell at a distance of 3.25 µm from the center?
Transcribed Image Text:Red blood cells can be modeled as spheres of 7.03 µm diameter with -2.55 × 10-12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space eo is 8.85 × 10-12 C/(V-m). What is the electric field Éj inside the cell at a distance of 3.25 µm from the center?
What is the electric field É, at a point outside the cell 3.25 µm from the surface? V/m
Transcribed Image Text:What is the electric field É, at a point outside the cell 3.25 µm from the surface? V/m
Expert Solution
Step 1

The electric field can be calculated with the help of Gauss' law of electrostatic.

Let us assume a symmetrical Gaussian surface at a distance r from the center of the sphere and outside the sphere. The Gauss' law for such a surface can be written as,

∫0rEds=Qε0.........................(1)

Here, E is the electric field, ds is any elemental surface area on the imaginary Gaussian surface of radius r, Q is the total charge enclosed by the Gaussian surface and ε0 is the permittivity of free space.

Here the direction of the electric field and the elemental surface area are considered to be positive along the direction away from the center of the sphere. That's why the dot product is written by considering the cosine term to be 1.

In the given problem, the Gaussian surface is outside the original sphere, which is the modeled red blood cell. Thus, total charge enclosed by the Gaussian surface is the total charge that lies on the surface of the red blood cell.

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