It takes work to push charge through a change of potential.
There's no change of potential along an equipotential path,
so that path doesn't require any work.
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.
Answer:
12552 J or 3000 calories
Explanation:
Q = m × c × ∆T
Where;
Q = amount of heat energy (J)
m = mass of water (g)
c = specific heat capacity (4.184 J/g°C)
∆T = change in temperature
For 50mL of water, there are 50g, hence, m = 50g, c = 4.184 J/g°C, initial temperature = 0°C, final temperature = 60°C.
Q = m × c × ∆T
Q = 50 × 4.184 × (60 - 0)
Q = 209.2 × 60
Q = 12552 J
Hence, the amount of heat energy used to heat the water is 12552 J or 3000 calories
<span>C. They decay at a predictable rate. IE carbon 14 decay to carbon 12 to understand how long ago something was alive.</span>
