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lukranit [14]
2 years ago
12

The electric field is measured for points at distances r from the center of a uniformly charged insulating sphere that has volum

e charge densi
Physics
1 answer:
RoseWind [281]2 years ago
4 0

For the electric field is measured for points at distances r,Electric field  is mathematically given as

E=19.9*10^{-5}c/m^3

<h3>What is the Electric field?</h3>

Generally, the equation for the  electric field uniform charged is mathematically given as

E=\frac{lr}{3e0}

Therefore

E=lr/3E0=3Ee0/r

Therefore

E=3*6*10^4*8.85*10^{-12}

E=19.9*10^{-5}c/m^3

In conclusion, E=19.9*10^{-5}c/m^3

E=19.9*10^{-5}c/m^3

Read more about Electric field

brainly.com/question/9383604

CQ

Complete question attached below

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Initilal velocity of projectile at an angle of 22.5° is 10m/s. Then what is the magnitude of range of projectile?
crimeas [40]

Answer:

0.707m

Explanation:

from formula of range i.e R=Usin2Q/g

3 0
3 years ago
How many electrons are in the outer energy level of an atom of carbon
xz_007 [3.2K]

Answer:

4 electrons

Explanation:

Carbon, is the group 14 element, with four electrons in its outer shell. Carbon always shares electrons to reach a complete valence shell, making bonds with other atoms.

8 0
3 years ago
Can someone help me
Illusion [34]
Thomas Edison is the answer im 100% sure of it.
5 0
3 years ago
A point charge of 5.7 µc moves at 4.5 × 105 m/s in a magnetic field that has a field strength of 3.2 mt, as shown in the diagram
Sauron [17]

The magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

<h3>What is magnetic force?</h3>

Magnetic force, attraction or repulsion that arises between electrically charged particles because of their motion.The magnitude of the magnetic force acting on the charge is given by:

\rm F=qvBsin\theta

where

The magnitude of the charge   q=5.7\ \mu C =5.7\times 10^{-6} \ C

The velocity of the charge        v=4.5\times 10^5\ \frac{m}{s}

The magnitude of the magnetic field   3.2\ mT=0.0032\ T

The angle between the directions of v and B  \theta =90^o-37^o=53^o

By substituting the values we will get:

F=(5.7\times 10^}-6})(4.5\times 10^5)(0.0032)(Sin53^o)

F=6.6\times 10^{-3}\ N

Hence the magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

To know more about Magnetic force follow

brainly.com/question/14411049

3 0
2 years ago
How much charge is on each plate of a 3.00-μF capacitor when it is connected toa 15.0-V battery? b) If this same capacitor is c
Sauron [17]

Answer:

(a) 45 micro coulomb

(b) 6 micro Coulomb

Explanation:

C = 3 micro Farad = 3 x 10^-6 Farad

V = 15 V

(a) q = C x V

where, q be the charge.

q = 3 x 10^-6 x 15 = 45 x 10^-6 C = 45 micro coulomb

(b)

V = 2 V, C = 3 micro Farad = 3 x 10^-6 Farad

q = C x V

where, q be the charge.

q = 3 x 10^-6 x 2 = 6 x 10^-6 C = 6 micro coulomb

6 0
3 years ago
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