Asolid sphere of radius 40.0 cm has a total positive charge of 17.4 µc uniformly distributed throughout its volume. calculate the magnitude of the electric field at the following distances. (a) 0 cm from the center of the sphere
kn/c

(b) 10.0 cm from the center of the sphere
/c

(c) 40.0 cm from the center of the sphere
kn/c (d) 71.0 cm from the center of the sphere
kn/c

(a)Magnitude of the electric field at  0 cm from the center of the sphere:

E=0              

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere :

  E= 244.7 KN/C

(c) Magnitude of the electric field at 40.0 cm from the center of the sphere :

  E=978.8 KN/C

(d) Magnitude of the electric field at 71 cm from the center of the sphere :  

   E=310.7 KN/C

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere:

 : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

 :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q:  Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=17.4 μC=17.4 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at  0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

E= 244.7 KN/C

(c) Magnitude of the electric field at 40.0 cm from the center of the sphere

r=a, We apply the Formula (1) :

E=978.8 KN/C

(d) Magnitude of the electric field at 71 cm from the center of the sphere  

r>a , We apply the Formula (1) :

E=310.7 KN/C

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.