Answer:
a force that pushes down on things
The electric field due to a point charge of 20uC at a distance of 1 meter away from it is 180000
.
First, you have to know that the space surrounding a load suffers some kind of disturbance, since a load located in that space will suffer a force. The disturbance that this charge creates around it is called an electric field.
In other words, an electric field exists in a certain region of space if, when introducing a charge called witness charge or test charge, it undergoes the action of an electric force.
The electric field E created by the point charge q at any point P, located at a distance r, is defined as:

where K is the constant of Coulomb's law.
In this case, you know:
- K= 9×10⁹

- q= 20 uC=20×10⁻⁶ C
- r= 1 m
Replacing in the definition of electric field:

Solving:
<u><em>E=180000 </em></u>
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Finally, the electric field due to a point charge of 20uC at a distance of 1 meter away from it is 180000
.
Learn more:
Answer:
Explanation:
q = 2e = 3.2 x 10^-19 C
mass, m = 6.68 x 10^-27 kg
Kinetic energy, K = 22 MeV
Current, i = 0.27 micro Ampere = 0.27 x 10^-6 A
(a) time, t = 2.8 s
Let N be the alpha particles strike the surface.
N x 2e = q
N x 3.2 x 10^-19 = i t
N x 3.2 x 10^-19 = 0.27 x 10^-6 x 2.8
N = 2.36 x 10^12
(b) Length, L = 16 cm = 0.16 m
Let N be the alpha particles
K = 0.5 x mv²
22 x 1.6 x 10^-13 = 0.5 x 6.68 x 10^-27 x v²
v² = 1.054 x 10^15
v = 3.25 x 10^7 m/s
So, N x 2e = i x t
N x 2e = i x L / v
N x 3.2 x 10^-19 = 2.7 x 10^-7 x 0.16 / (3.25 x 10^7)
N = 4153.85
(c) Us ethe conservation of energy
Kinetic energy = Potential energy
K = q x V
22 x 1.6 x 10^-13 = 2 x 1.5 x 10^-19 x V
V = 1.17 x 10^7 V
Answer:
B
Explanation:
because. it reduces the distance between the earth