Answer with explanation :
The negative sign means that the potential energy decreases by the movement of the electron.
negative charge at rest in an electric field moves toward the region of an electric field , so that its potential energy will diminish and change into the kinetic energy of motion. The total energy remains constant.
Positive charges will move downhill because of convention. It is to stay in accordance with other potential theories, particularly gravity, where the "charge" is mass, that moves downwards in the gravitational potential field expressed by ϕ(r)=−GM|r|ϕ(r)=−GM|r|. In an electronic system, howbeit, positive charges are fixed in their position within a component (e.g., a wire), therefore instead of the mobile,the negative charges, electrons, move uphill.
Answer:
0.84 m
Explanation:
Given in the y direction:
Δy = 0.60 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
0.60 m = (0 m/s) t + ½ (9.8 m/s²) t²
t = 0.35 s
Given in the x direction:
v₀ = 2.4 m/s
a = 0 m/s²
t = 0.35 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (2.4 m/s) (0.35 s) + ½ (0 m/s²) (0.35 s)²
Δx = 0.84 m
Not sure.can you give me a clue?
Answer:
1.38 x 10^-18 J
Explanation:
q = - 1.6 x 10^-19 C
d = 5 x 10^-10 m
the potential energy of the system gives the value of work done
The formula for the potential energy is given by
So, the total potential energy of teh system is
As all the charges are same and the distance between the two charges is same so the total potential energy becomes
K = 9 x 10^9 Nm^2/C^2
By substituting the values
U = 1.38 x 10^-18 J
Answer:
Same magnitude of the 10 nc charge cause the electric field is external.
Explanation:
To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.
As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.
F = qE
If charge is doubled, then the force would be bigger in the second case than in the first case, but electric field remain the same value.
