Answer:
the branch of science concerned with the nature and properties of matter and energy
Answer:
V = 192 kV
Explanation:
Given that,
Charge, 
Distance, r = 0.3 m
We need to find the electric potential at a distance of 0.3 m from a point charge. The formula for electric potential is given by :

So, the required electric potential is 192 kV.
Answer:
Wavelength of radio is wave is 3 m
Explanation:
Wavelength of radio is wave is

where

wavelength is

T<span>he equation to be used here to determine the distance between two equipotential points is:
V = k * Q / r
where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance.
In this case, we can use ratio of proportions to determine the distance between the two points. in this respect,
Point 1:
V = k * Q / r = 290
r = k*Q/290 ; kQ = 290r
Point 2:
V = k * Q / R = 41
R = k*Q/41
from equation 10 kQ = 290r
R = 290/(41)= 7.07 m
The distance between the two points then is equal to 7.07 m.
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