The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m
Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:
Substitute numerical values:
The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
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Answer:
3.6 x 10⁶ Pa
Explanation:
A = Area of the heel = 1.50 cm² = 1.50 x 10⁻⁴ m²
m = mass of the woman = 55.0 kg
g = acceleration due to gravity = 9.8 m/s²
Force of gravity on the heel is given as
F = mg
Inserting the values
F = (55) (9.8)
F = 539 N
Pressure exerted on the floor is given as
P = 3.6 x 10⁶ Pa